Explanation:
Okay, so there's this pirate who's searching seven islands for buried treasure. Each island has a 1/5 chance of having treasure. The question is asking for the probability that exactly 4 of the islands have treasure. Hmm, let me think about how to approach this.

First, I remember that when dealing with probabilities of a certain number of successes in a fixed number of independent trials, we can use the binomial probability formula. Is this a binomial situation? Let me check the conditions. There are a fixed number of trials, which here are the seven islands. Each trial (island) has two possible outcomes: treasure or no treasure. The probability of success (finding treasure) is constant at 1/5 for each island. And the trials are independent because the presence of treasure on one island doesn't affect the others. So yes, this is a binomial distribution problem.

The formula for binomial probability is P(k) = C(n, k) * p^k * (1 - p)^(n - k), where:
- P(k) is the probability of exactly k successes,
- C(n, k) is the combination of n things taken k at a time,
- p is the probability of success on a single trial,
- n is the number of trials.

In this case, n = 7, k = 4, p = 1/5. So plugging these into the formula, we get:

P(4) = C(7, 4) * (1/5)^4 * (4/5)^(7 - 4)

Let me compute each part step by step.

First, calculate C(7, 4). The combination formula is C(n, k) = n! / (k! * (n - k)!).

So C(7, 4) = 7! / (4! * (7 - 4)!) = 7! / (4! * 3!)

Calculating factorials:
7! = 5040
4! = 24
3! = 6

So C(7, 4) = 5040 / (24 * 6) = 5040 / 144. Let me compute that. Dividing 5040 by 144. Let's see, 144 * 35 = 5040 because 144 * 30 = 4320, and 144 * 5 = 720, so 4320 + 720 = 5040. So C(7, 4) = 35.

Okay, so that part is 35.

Next, (1/5)^4. Let me compute that. (1/5)^4 = 1 / (5^4) = 1 / 625.

Then, (4/5)^(7 - 4) = (4/5)^3. Let me calculate that. (4/5)^3 = 64 / 125.

So putting it all together:

P(4) = 35 * (1/625) * (64/125)

Now, let me compute this step by step. First, multiply 35 and 1/625. 35 / 625. Let me simplify that fraction. Both numerator and denominator are divisible by 5: 35 ÷ 5 = 7, 625 ÷ 5 = 125. So that becomes 7/125.

Then, multiply that by 64/125. So 7/125 * 64/125. Multiply numerators: 7 * 64 = 448. Multiply denominators: 125 * 125 = 15625. So the probability is 448 / 15625.

Wait, let me check if that can be simplified. Let's see if 448 and 15625 have any common factors. 448 factors into 2^6 * 7 (since 448 ÷ 2 = 224, ÷2=112, ÷2=56, ÷2=28, ÷2=14, ÷2=7). 15625 is 5^6. There are no common prime factors between numerator and denominator, so 448/15625 is the simplified fraction.

Alternatively, if I want to write it as a decimal, I can divide 448 by 15625. Let me do that. Let me see:

First, note that 15625 * 0.028 = 15625 * 0.02 + 15625 * 0.008 = 312.5 + 125 = 437.5. Hmm, 0.028 gives 437.5. The numerator is 448, which is 448 - 437.5 = 10.5 more. So 10.5 / 15625 = 0.000672. So total is approximately 0.028 + 0.000672 = 0.028672. So approximately 2.8672%.

But the question doesn't specify the form of the answer, so maybe they want the exact fraction. Let me check if I did all the calculations correctly.

Wait, let me verify the combination again. C(7,4) is 35. Yes, because C(7,4) is same as C(7,3) which is 35. Let me confirm with another method. For example, 7 choose 4: 7*6*5*4 / 4*3*2*1 = (7*6*5*4)/24. Let me compute that. 7*6=42, 42*5=210, 210*4=840. 840 divided by 24 is 35. Yes, that's correct.

Then (1/5)^4 is 1/625, and (4/5)^3 is 64/125. Multiplying them together with 35: 35*(1/625)*(64/125). Let me check the multiplication again. 35*64 = 2240, and then divided by 625*125 = 78125? Wait, wait, no. Wait, 35*(1/625)*(64/125) is equal to (35*64)/(625*125). Let me compute denominator: 625*125. 625 is 5^4, 125 is 5^3, so 5^7 = 78125. Numerator: 35*64. Let me compute 35*64. 35*60=2100, 35*4=140, so total is 2100+140=2240. So 2240 / 78125. Wait, earlier I had 448 / 15625. Wait, 2240 divided by 5 is 448, and 78125 divided by 5 is 15625. So yes, 2240/78125 = 448/15625. So that's correct.

So the exact probability is 448/15625. Let me check if that reduces further. As I thought before, 448 is 2^6 * 7, and 15625 is 5^6. No common factors, so that's the simplified fraction. Alternatively, if they want a decimal, approximately 0.028672, which is about 2.87%.

But let me check if I made a mistake in the exponents. Wait, the formula is C(n, k) * p^k * (1 - p)^{n - k}. So n is 7, k is 4, so (1 - p) is 4/5, and the exponent is 7 - 4 = 3. So that's correct. So (4/5)^3. So that's correct. So the calculation seems right.

Alternatively, maybe I can think of it as 35 * (1/5)^4 * (4/5)^3. Let me compute that again. 35*(1/5)^4*(4/5)^3. Let me write all terms with denominator 5^7. Because (1/5)^4 is 1/5^4 and (4/5)^3 is 4^3 / 5^3. So multiplying them together, we have 1*4^3 / 5^(4+3) = 64 / 5^7. Then multiplied by 35. So total is 35*64 / 5^7. Let me compute 5^7. 5^2=25, 5^3=125, 5^4=625, 5^5=3125, 5^6=15625, 5^7=78125. So yes, denominator is 78125. Numerator is 35*64. Which is 2240. So 2240/78125. Which reduces to 448/15625. Yep, same as before.

So that seems correct. Therefore, the probability is 448/15625. If I want to write that as a decimal, as I did before, approximately 0.028672, which is about 2.87%. But maybe the question expects the fractional form. Let me check if 448/15625 can be simplified further. Let me check GCD of 448 and 15625. The prime factors of 448 are 2^6 * 7. The prime factors of 15625 are 5^6. There are no common prime factors, so yes, it's already in simplest terms.

Alternatively, maybe I can write it as a decimal or a percentage, but unless specified, the fraction is probably the better answer. Let me check if 448 divided by 15625 is indeed the correct value. Let me do a quick check with another approach.

Alternatively, think of the probability as the number of ways to choose 4 islands out of 7, times the probability that those 4 have treasure and the remaining 3 don't. So that's exactly what we did. So 35 ways, each with probability (1/5)^4*(4/5)^3. So that's correct.

Alternatively, maybe I can use decimal approximations step by step to verify. Let me compute 35 * (1/5)^4 * (4/5)^3.

First, compute (1/5)^4: 1/5 is 0.2, so 0.2^4 = 0.0016.

Then, (4/5)^3: 4/5 is 0.8, so 0.8^3 = 0.512.

Multiply those two: 0.0016 * 0.512 = 0.0008192.

Then multiply by 35: 0.0008192 * 35. Let me compute that. 0.0008192 * 35. Let me think of 0.0008192 * 35 = 0.0008192 * 30 + 0.0008192 * 5 = 0.024576 + 0.004096 = 0.028672. Which matches the decimal I had earlier. So 0.028672, which is 448/15625. So that seems correct.

Therefore, the probability is 448/15625. If I want to write that as a reduced fraction, but since it's already reduced, that's the answer. Alternatively, maybe the problem expects the answer in some other form? Let me check if 448/15625 can be simplified. Wait, 448 divided by... Let me check if 448 and 15625 have any common divisors. 448 is even, 15625 is odd, so no factor of 2. 448 divided by 7 is 64, and 15625 divided by 7? Let me check 7*2232 = 15624, so 15625 is 7*2232 +1, so not divisible by 7. Therefore, no. So yes, 448/15625 is simplest.

Alternatively, maybe the problem expects the answer as a decimal or percentage? But since it's a math problem, and unless specified, fractions are usually preferred. So I think 448/15625 is the answer. Let me check if that's the case.

Alternatively, maybe I made a mistake in the combination? Let me check again. C(7,4). 7 choose 4. Let me think of it as 7*6*5*4 / 4*3*2*1. Which is (7*6*5*4)/24. Let me compute numerator: 7*6 is 42, 42*5 is 210, 210*4 is 840. 840 divided by 24. 24*35 is 840. So yes, 35. Correct.

Alternatively, maybe I miscalculated the exponents? Let me check again. The formula is C(n,k) * p^k * (1-p)^{n -k}. So n=7, k=4, so (1-p)^{7 -4} = (4/5)^3. Correct. And p^k is (1/5)^4. Correct. So that's right.

Therefore, I think my answer is correct. So the probability is 448/15625. If I want to write that as a decimal, approximately 0.028672, which is about 2.87%. But since the question doesn't specify, I think the exact fraction is better.

Wait, but let me check once more with another approach. Suppose I list all possible ways. But with seven islands, that's too many. Alternatively, maybe use the binomial probability formula in another way. But I think the method is correct.

Alternatively, maybe I can use logarithms or something else? Probably not necessary. The calculation seems straightforward.

Alternatively, maybe I can check with a calculator? Well, since I don't have a calculator here, but I can verify the decimal value. Let me compute 448 divided by 15625. Let me do that division step by step.

First, note that 15625 x 0.02 = 312.5

Subtract that from 448: 448 - 312.5 = 135.5

Now, 15625 x 0.008 = 125. So 0.02 + 0.008 = 0.028, and 135.5 - 125 = 10.5

Now, 15625 x 0.0006 = 9.375

So adding 0.0006 to 0.028 gives 0.0286, and 10.5 - 9.375 = 1.125

Then, 15625 x 0.00007 = approx 1.09375

Adding 0.00007 to 0.0286 gives 0.02867, and 1.125 - 1.09375 = 0.03125

Then, 15625 x 0.000002 = 0.03125

So adding 0.000002 to 0.02867 gives 0.028672. So total is 0.028672, which matches the earlier decimal. Therefore, 448/15625 is indeed approximately 0.028672.

Therefore, I think my answer is correct. So the probability that exactly 4 of the islands have treasure is 448/15625.

**Final Answer**
The probability that exactly 4 of the islands have treasure is \boxed{\dfrac{448}{15625}}.

To determine the probability that exactly 4 out of 7 islands have treasure, we model this as a **binomial probability** problem. Here's the step-by-step breakdown:

---

### **Step 1: Identify the Parameters**

- **Number of trials (n):** 7 (since there are 7 islands)
- **Number of successes (k):** 4 (we want exactly 4 islands with treasure)
- **Probability of success (p):** $ \frac{1}{5} $ (each island has a 1/5 chance of having treasure)
- **Probability of failure (1 - p):** $ \frac{4}{5} $

---

### **Step 2: Apply the Binomial Probability Formula**

The binomial probability formula is:

$$
P(k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}
$$

Substituting the known values:

$$
P(4) = \binom{7}{4} \cdot \left(\frac{1}{5}\right)^4 \cdot \left(\frac{4}{5}\right)^3
$$

---

### **Step 3: Compute Each Component**

- **Combination:**
  $$
  \binom{7}{4} = \frac{7!}{4!(7 - 4)!} = \frac{7 \cdot 6 \cdot 5 \cdot 4}{4 \cdot 3 \cdot 2 \cdot 1} = 35
  $$

- **Probability of 4 successes:**
  $$
  \left(\frac{1}{5}\right)^4 = \frac{1}{625}
  $$

- **Probability of 3 failures:**
  $$
  \left(\frac{4}{5}\right)^3 = \frac{64}{125}
  $$

---

### **Step 4: Multiply All Components**

$$
P(4) = 35 \cdot \frac{1}{625} \cdot \frac{64}{125}
$$

First, multiply the numerators:

$$
35 \cdot 64 = 2240
$$

Then, multiply the denominators:

$$
625 \cdot 125 = 78125
$$

So:

$$
P(4) = \frac{2240}{78125}
$$

Simplify the fraction by dividing numerator and denominator by 5:

$$
\frac{2240}{78125} = \frac{448}{15625}
$$

---

### **Final Answer**

The probability that exactly 4 of the islands have treasure is:

$$
\boxed{\dfrac{448}{15625}}
$$

[Assistant]
Explanation:
Okay, so there's this pirate who's searching seven islands for buried treasure. Each island has a 1/5 chance of having treasure. The question is asking for the probability that exactly 4 of the islands have treasure. Hmm, let me think about how to approach this.

First, I remember that when dealing with probabilities of a certain number of successes in a fixed number of independent trials, we can use the binomial probability formula. Is this a binomial situation? Let me check the conditions. There are a fixed number of trials, which here are the seven islands. Each trial (island) has two possible outcomes: treasure or no treasure. The probability of success (finding treasure) is constant at 1/5 for each island. And the trials are independent because the presence of treasure on one island doesn't affect the others. So yes, this is a binomial distribution problem.

The formula for binomial probability is P(k) = C(n, k) * p^k * (1 - p)^(n - k), where:
- P(k) is the probability of exactly k successes,
- C(n, k) is the combination of n things taken k at a time,
- p is the probability of success on a single trial,
- n is the number of trials.

In this case, n = 7, k = 4, p = 1/5. So plugging these into the formula, we get:

P(4) = C(7, 4) * (1/5)^4 * (4/5)^(7 - 4)

Let me compute each part step by step.

First, calculate C(7, 4). The combination formula is C(n, k) = n! / (k! * (n - k)!).

So C(7, 4) = 7! / (4! * (7 - 4)!) = 7! / (4! * 3!)

Calculating factorials:
7! = 5040
4! = 24
3! = 6

So C(7, 4) = 5040 / (24 * 6) = 5040 / 144. Let me compute that. Dividing 5040 by 144. Let's see, 144 * 35 = 5040 because 144 * 30 = 4320, and 144 * 5 = 720, so 4320 + 720 = 5040. So C(7, 4) = 35.

Okay, so that part is 35.

Next, (1/5)^4. Let me compute that. (1/5)^4 = 1 / (5^4) = 1 / 625.

Then, (4/5)^(7 - 4) = (4/5)^3. Let me calculate that. (4/5)^3 = 64 / 125.

So putting it all together:

P(4) = 35 * (1/625) * (64/125)

Now, let me compute this step by step. First, multiply 35 and 1/625. 35 / 625. Let me simplify that fraction. Both numerator and denominator are divisible by 5: 35 ÷ 5 = 7, 625 ÷ 5 = 125. So that becomes 7/125.

Then, multiply that by 64/125. So 7/125 * 64/125. Multiply numerators: 7 * 64 = 448. Multiply denominators: 125 * 125 = 15625. So the probability is 448 / 15625.

Wait, let me check if that can be simplified. Let's see if 448 and 15625 have any common factors. 448 factors into 2^6 * 7 (since 448 ÷ 2 = 224, ÷2=112, ÷2=56, ÷2=28, ÷2=14, ÷2=7). 15625 is 5^6. There are no common prime factors between numerator and denominator, so 448/15625 is the simplified fraction.

Alternatively, if I want to write it as a decimal, I can divide 448 by 15625. Let me do that. Let me see:

First, note that 15625 * 0.028 = 15625 * 0.02 + 15625 * 0.008 = 312.5 + 125 = 437.5. Hmm, 0.028 gives 437.5. The numerator is 448, which is 448 - 437.5 = 10.5 more. So 10.5 / 15625 = 0.000672. So total is approximately 0.028 + 0.000672 = 0.028672. So approximately 2.8672%.

But the question doesn't specify the form of the answer, so maybe they want the exact fraction. Let me check if I did all the calculations correctly.

Wait, let me verify the combination again. C(7,4) is 35. Yes, because C(7,4) is same as C(7,3) which is 35. Let me confirm with another method. For example, 7 choose 4: 7*6*5*4 / 4*3*2*1 = (7*6*5*4)/24. Let me compute that. 7*6=42, 42*5=210, 210*4=840. 840 divided by 24 is 35. Yes, that's correct.

Then (1/5)^4 is 1/625, and (4/5)^3 is 64/125. Multiplying them together with 35: 35*(1/625)*(64/125). Let me check the multiplication again. 35*64 = 2240, and then divided by 625*125 = 78125? Wait, wait, no. Wait, 35*(1/625)*(64/125) is equal to (35*64)/(625*125). Let me compute denominator: 625*125. 625 is 5^4, 125 is 5^3, so 5^7 = 78125. Numerator: 35*64. Let me compute 35*64. 35*60=2100, 35*4=140, so total is 2100+140=2240. So 2240 / 78125. Wait, earlier I had 448 / 15625. Wait, 2240 divided by 5 is 448, and 78125 divided by 5 is 15625. So yes, 2240/78125 = 448/15625. So that's correct.

So the exact probability is 448/15625. Let me check if that reduces further. As I thought before, 448 is 2^6 * 7, and 15625 is 5^6. No common factors, so that's the simplified fraction. Alternatively, if they want a decimal, approximately 0.028672, which is about 2.87%.

But let me check if I made a mistake in the exponents. Wait, the formula is C(n, k) * p^k * (1 - p)^{n - k}. So n is 7, k is 4, so (1 - p) is 4/5, and the exponent is 7 - 4 = 3. So that's correct. So (4/5)^3. So that's correct. So the calculation seems right.

Alternatively, maybe I can think of it as 35 * (1/5)^4 * (4/5)^3. Let me compute that again. 35*(1/5)^4*(4/5)^3. Let me write all terms with denominator 5^7. Because (1/5)^4 is 1/5^4 and (4/5)^3 is 4^3 / 5^3. So multiplying them together, we have 1*4^3 / 5^(4+3) = 64 / 5^7. Then multiplied by 35. So total is 35*64 / 5^7. Let me compute 5^7. 5^2=25, 5^3=125, 5^4=625, 5^5=3125, 5^6=15625, 5^7=78125. So yes, denominator is 78125. Numerator is 35*64. Which is 2240. So 2240/78125. Which reduces to 448/15625. Yep, same as before.

So that seems correct. Therefore, the probability is 448/15625. If I want to write that as a decimal, as I did before, approximately 0.028672, which is about 2.87%. But maybe the question expects the fractional form. Let me check if 448/15625 can be simplified further. Let me check GCD of 448 and 15625. The prime factors of 448 are 2^6 * 7. The prime factors of 15625 are 5^6. There are no common prime factors, so yes, it's already in simplest terms.

Alternatively, maybe I can write it as a decimal or a percentage, but unless specified, the fraction is probably the better answer. Let me check if 448 divided by 15625 is indeed the correct value. Let me do a quick check with another approach.

Alternatively, think of the probability as the number of ways to choose 4 islands out of 7, times the probability that those 4 have treasure and the remaining 3 don't. So that's exactly what we did. So 35 ways, each with probability (1/5)^4*(4/5)^3. So that's correct.

Alternatively, maybe I can use decimal approximations step by step to verify. Let me compute 35 * (1/5)^4 * (4/5)^3.

First, compute (1/5)^4: 1/5 is 0.2, so 0.2^4 = 0.0016.

Then, (4/5)^3: 4/5 is 0.8, so 0.8^3 = 0.512.

Multiply those two: 0.0016 * 0.512 = 0.0008192.

Then multiply by 35: 0.0008192 * 35. Let me compute that. 0.0008192 * 35. Let me think of 0.0008192 * 35 = 0.0008192 * 30 + 0.0008192 * 5 = 0.024576 + 0.004096 = 0.028672. Which matches the decimal I had earlier. So 0.028672, which is 448/15625. So that seems correct.

Therefore, the probability is 448/15625. If I want to write that as a reduced fraction, but since it's already reduced, that's the answer. Alternatively, maybe the problem expects the answer in some other form? Let me check if 448/15625 can be simplified. Wait, 448 divided by... Let me check if 448 and 15625 have any common divisors. 448 is even, 15625 is odd, so no factor of 2. 448 divided by 7 is 64, and 15625 divided by 7? Let me check 7*2232 = 15624, so 15625 is 7*2232 +1, so not divisible by 7. Therefore, no. So yes, 448/15625 is simplest.

Alternatively, maybe the problem expects the answer as a decimal or percentage? But since it's a math problem, and unless specified, fractions are usually preferred. So I think 448/15625 is the answer. Let me check if that's the case.

Alternatively, maybe I made a mistake in the combination? Let me check again. C(7,4). 7 choose 4. Let me think of it as 7*6*5*4 / 4*3*2*1. Which is (7*6*5*4)/24. Let me compute numerator: 7*6 is 42, 42*5 is 210, 210*4 is 840. 840 divided by 24. 24*35 is 840. So yes, 35. Correct.

Alternatively, maybe I miscalculated the exponents? Let me check again. The formula is C(n,k) * p^k * (1-p)^{n -k}. So n=7, k=4, so (1-p)^{7 -4} = (4/5)^3. Correct. And p^k is (1/5)^4. Correct. So that's right.

Therefore, I think my answer is correct. So the probability is 448/15625. If I want to write that as a decimal, approximately 0.028672, which is about 2.87%. But since the question doesn't specify, I think the exact fraction is better.

Wait, but let me check once more with another approach. Suppose I list all possible ways. But with seven islands, that's too many. Alternatively, maybe use the binomial probability formula in another way. But I think the method is correct.

Alternatively, maybe I can use logarithms or something else? Probably not necessary. The calculation seems straightforward.

Alternatively, maybe I can check with a calculator? Well, since I don't have a calculator here, but I can verify the decimal value. Let me compute 448 divided by 15625. Let me do that division step by step.

First, note that 15625 x 0.02 = 312.5

Subtract that from 448: 448 - 312.5 = 135.5

Now, 15625 x 0.008 = 125. So 0.02 + 0.008 = 0.028, and 135.5 - 125 = 10.5

Now, 15625 x 0.0006 = 9.375

So adding 0.0006 to 0.028 gives 0.0286, and 10.5 - 9.375 = 1.125

Then, 15625 x 0.00007 = approx 1.09375

Adding 0.00007 to 0.0286 gives 0.02867, and 1.125 - 1.09375 = 0.03125

Then, 15625 x 0.000002 = 0.03125

So adding 0.000002 to 0.02867 gives 0.028672. So total is 0.028672, which matches the earlier decimal. Therefore, 448/15625 is indeed approximately 0.028672.

Therefore, I think my answer is correct. So the probability that exactly 4 of the islands have treasure is 448/15625.

**Final Answer**
The probability that exactly 4 of the islands have treasure is \boxed{\dfrac{448}{15625}}.

To determine the probability that exactly 4 out of 7 islands have treasure, we model this as a **binomial probability** problem. Here's the step-by-step breakdown:

---

### **Step 1: Identify the Parameters**

- **Number of trials (n):** 7 (since there are 7 islands)
- **Number of successes (k):** 4 (we want exactly 4 islands with treasure)
- **Probability of success (p):** $ \frac{1}{5} $ (each island has a 1/5 chance of having treasure)
- **Probability of failure (1 - p):** $ \frac{4}{5} $

---

### **Step 2: Apply the Binomial Probability Formula**

The binomial probability formula is:

$$
P(k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}
$$

Substituting the known values:

$$
P(4) = \binom{7}{4} \cdot \left(\frac{1}{5}\right)^4 \cdot \left(\frac{4}{5}\right)^3
$$

---

### **Step 3: Compute Each Component**

- **Combination:**
  $$
  \binom{7}{4} = \frac{7!}{4!(7 - 4)!} = \frac{7 \cdot 6 \cdot 5 \cdot 4}{4 \cdot 3 \cdot 2 \cdot 1} = 35
  $$

- **Probability of 4 successes:**
  $$
  \left(\frac{1}{5}\right)^4 = \frac{1}{625}
  $$

- **Probability of 3 failures:**
  $$
  \left(\frac{4}{5}\right)^3 = \frac{64}{125}
  $$

---

### **Step 4: Multiply All Components**

$$
P(4) = 35 \cdot \frac{1}{625} \cdot \frac{64}{125}
$$

First, multiply the numerators:

$$
35 \cdot 64 = 2240
$$

Then, multiply the denominators:

$$
625 \cdot 125 = 78125
$$

So:

$$
P(4) = \frac{2240}{78125}
$$

Simplify the fraction by dividing numerator and denominator by 5:

$$
\frac{2240}{78125} = \frac{448}{15625}
$$

---

### **Final Answer**

The probability that exactly 4 of the islands have treasure is:

$$
\boxed{\dfrac{448}{15625}}
$$

Solution: There are $\binom{7}{4}=35$ ways to choose 4 of the islands. For each choice, there is a probability of $\left( \frac{1}{5} \right)^4 \left( \frac{4}{5} \right)^3$ that those 4 islands will have treasure and the others will not. Therefore, the total probability that exactly 4 of the islands have treasure is $35 \left( \frac{1}{5} \right)^4 \left( \frac{4}{5} \right)^3 = \boxed{\frac{448}{15625}}$.


Explanation:
Okay, so I need to find the value of $ c $ such that the circle given by the equation $ x^2 - 10x + y^2 + 6y + c = 0 $ has a radius of length 1. Hmm, let me recall how to approach this.

First, I remember that the general equation of a circle is $ (x - h)^2 + (y - k)^2 = r^2 $, where $ (h, k) $ is the center and $ r $ is the radius. The given equation isn't in that standard form, so I probably need to complete the square for both the $ x $ and $ y $ terms. That should help me rewrite the equation into the standard form and then compare it to find the value of $ c $.

Let me start by rearranging the given equation. The original equation is:

$ x^2 - 10x + y^2 + 6y + c = 0 $

I need to group the $ x $ terms and the $ y $ terms together. Let me rewrite it as:

$ (x^2 - 10x) + (y^2 + 6y) + c = 0 $

Now, I need to complete the square for each of the quadratic terms. Let me handle the $ x $ terms first: $ x^2 - 10x $. To complete the square, I take the coefficient of $ x $, which is -10, divide by 2 to get -5, and then square it to get 25. So, I can write:

$ x^2 - 10x = (x - 5)^2 - 25 $

Wait, let me check that. If I expand $ (x - 5)^2 $, it's $ x^2 - 10x + 25 $, so to get back to $ x^2 - 10x $, I need to subtract 25. So yes, that's correct. So, $ x^2 - 10x = (x - 5)^2 - 25 $.

Now for the $ y $ terms: $ y^2 + 6y $. Similarly, take the coefficient of $ y $, which is 6, divide by 2 to get 3, square it to get 9. Therefore:

$ y^2 + 6y = (y + 3)^2 - 9 $

Let me verify that. Expanding $ (y + 3)^2 $ gives $ y^2 + 6y + 9 $, so subtracting 9 gives back $ y^2 + 6y $. Correct.

So substituting these back into the original equation:

$ (x - 5)^2 - 25 + (y + 3)^2 - 9 + c = 0 $

Now combine the constants:

$ (x - 5)^2 + (y + 3)^2 - 25 - 9 + c = 0 $

Which simplifies to:

$ (x - 5)^2 + (y + 3)^2 + (c - 34) = 0 $

Wait, hold on. Let me check the arithmetic again. The original equation after substitution is:

$ (x - 5)^2 - 25 + (y + 3)^2 - 9 + c = 0 $

So combining the constants: -25 -9 is -34, so:

$ (x - 5)^2 + (y + 3)^2 - 34 + c = 0 $

Therefore, moving the constants to the other side:

$ (x - 5)^2 + (y + 3)^2 = 34 - c $

Ah, right. So the standard form is $ (x - h)^2 + (y - k)^2 = r^2 $, so comparing, we have:

$ r^2 = 34 - c $

But the problem states that the radius should be 1. Therefore, $ r = 1 $, so $ r^2 = 1^2 = 1 $. Therefore:

$ 34 - c = 1 $

Solving for $ c $:

$ -c = 1 - 34 $

$ -c = -33 $

Multiply both sides by -1:

$ c = 33 $

Wait, so is that the answer? Let me check my steps again to make sure I didn't make a mistake.

Original equation: $ x^2 - 10x + y^2 + 6y + c = 0 $

Completing the square for x: $ x^2 -10x $ becomes $ (x - 5)^2 -25 $

For y: $ y^2 +6y $ becomes $ (y + 3)^2 -9 $

So substituting back:

$ (x - 5)^2 -25 + (y + 3)^2 -9 + c = 0 $

Combine constants: -25 -9 = -34, so:

$ (x - 5)^2 + (y + 3)^2 -34 + c = 0 $

Which rearranges to:

$ (x - 5)^2 + (y + 3)^2 = 34 - c $

Yes, that's correct. Then since radius squared is 1, so 34 - c = 1, so c = 34 -1 = 33. So c is 33.

Wait, but let me check if I didn't mix up any signs. Let me verify with another approach. Let me recall that the general equation $ x^2 + y^2 + Dx + Ey + F = 0 $ can be rewritten as $ (x + D/2)^2 + (y + E/2)^2 = (D/2)^2 + (E/2)^2 - F $. Therefore, the radius squared is $ (D/2)^2 + (E/2)^2 - F $.

In our case, the equation is $ x^2 -10x + y^2 +6y + c = 0 $. Comparing to the general form, D is -10, E is 6, and F is c. Therefore, radius squared is $ (-10/2)^2 + (6/2)^2 - c $, which is $ (-5)^2 + 3^2 - c = 25 + 9 - c = 34 - c $. Which matches what I had earlier. Therefore, radius squared is 34 - c. Since radius is 1, 34 - c = 1, so c = 33.

Therefore, the answer is 33. Let me check once more with plugging back in. Suppose c = 33, then the equation becomes:

$ x^2 -10x + y^2 +6y +33 = 0 $

Completing the square:

$ (x - 5)^2 -25 + (y + 3)^2 -9 +33 = 0 $

Which is $ (x - 5)^2 + (y + 3)^2 -25 -9 +33 = 0 $

Calculating constants: -25 -9 = -34, then -34 +33 = -1. Therefore:

$ (x - 5)^2 + (y + 3)^2 -1 = 0 $

Which is $ (x - 5)^2 + (y + 3)^2 = 1 $. So radius squared is 1, radius is 1. Perfect, that checks out.

Therefore, the value of c is indeed 33. I think that's correct. I don't see any mistakes in my reasoning. So I think the answer is 33.

**Final Answer**
The value of $ c $ is \boxed{33}.

To determine the value of $ c $ such that the circle defined by the equation

$$
x^2 - 10x + y^2 + 6y + c = 0
$$

has a radius of length 1, we begin by rewriting the equation in the standard form of a circle:

$$
(x - h)^2 + (y - k)^2 = r^2
$$

---

### Step 1: Complete the Square

We complete the square for both the $ x $ and $ y $ terms.

#### For the $ x $-terms:
$$
x^2 - 10x = (x - 5)^2 - 25
$$

#### For the $ y $-terms:
$$
y^2 + 6y = (y + 3)^2 - 9
$$

Substituting these into the original equation:

$$
(x - 5)^2 - 25 + (y + 3)^2 - 9 + c = 0
$$

Simplify the constants:

$$
(x - 5)^2 + (y + 3)^2 - 34 + c = 0
$$

Rearranging:

$$
(x - 5)^2 + (y + 3)^2 = 34 - c
$$

---

### Step 2: Use the Radius Condition

We are told the radius is 1, so the radius squared is:

$$
r^2 = 1^2 = 1
$$

From the equation above, the right-hand side is $ 34 - c $, so:

$$
34 - c = 1
$$

Solving for $ c $:

$$
c = 34 - 1 = 33
$$

---

### Final Answer

$$
\boxed{33}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ c $ such that the circle given by the equation $ x^2 - 10x + y^2 + 6y + c = 0 $ has a radius of length 1. Hmm, let me recall how to approach this.

First, I remember that the general equation of a circle is $ (x - h)^2 + (y - k)^2 = r^2 $, where $ (h, k) $ is the center and $ r $ is the radius. The given equation isn't in that standard form, so I probably need to complete the square for both the $ x $ and $ y $ terms. That should help me rewrite the equation into the standard form and then compare it to find the value of $ c $.

Let me start by rearranging the given equation. The original equation is:

$ x^2 - 10x + y^2 + 6y + c = 0 $

I need to group the $ x $ terms and the $ y $ terms together. Let me rewrite it as:

$ (x^2 - 10x) + (y^2 + 6y) + c = 0 $

Now, I need to complete the square for each of the quadratic terms. Let me handle the $ x $ terms first: $ x^2 - 10x $. To complete the square, I take the coefficient of $ x $, which is -10, divide by 2 to get -5, and then square it to get 25. So, I can write:

$ x^2 - 10x = (x - 5)^2 - 25 $

Wait, let me check that. If I expand $ (x - 5)^2 $, it's $ x^2 - 10x + 25 $, so to get back to $ x^2 - 10x $, I need to subtract 25. So yes, that's correct. So, $ x^2 - 10x = (x - 5)^2 - 25 $.

Now for the $ y $ terms: $ y^2 + 6y $. Similarly, take the coefficient of $ y $, which is 6, divide by 2 to get 3, square it to get 9. Therefore:

$ y^2 + 6y = (y + 3)^2 - 9 $

Let me verify that. Expanding $ (y + 3)^2 $ gives $ y^2 + 6y + 9 $, so subtracting 9 gives back $ y^2 + 6y $. Correct.

So substituting these back into the original equation:

$ (x - 5)^2 - 25 + (y + 3)^2 - 9 + c = 0 $

Now combine the constants:

$ (x - 5)^2 + (y + 3)^2 - 25 - 9 + c = 0 $

Which simplifies to:

$ (x - 5)^2 + (y + 3)^2 + (c - 34) = 0 $

Wait, hold on. Let me check the arithmetic again. The original equation after substitution is:

$ (x - 5)^2 - 25 + (y + 3)^2 - 9 + c = 0 $

So combining the constants: -25 -9 is -34, so:

$ (x - 5)^2 + (y + 3)^2 - 34 + c = 0 $

Therefore, moving the constants to the other side:

$ (x - 5)^2 + (y + 3)^2 = 34 - c $

Ah, right. So the standard form is $ (x - h)^2 + (y - k)^2 = r^2 $, so comparing, we have:

$ r^2 = 34 - c $

But the problem states that the radius should be 1. Therefore, $ r = 1 $, so $ r^2 = 1^2 = 1 $. Therefore:

$ 34 - c = 1 $

Solving for $ c $:

$ -c = 1 - 34 $

$ -c = -33 $

Multiply both sides by -1:

$ c = 33 $

Wait, so is that the answer? Let me check my steps again to make sure I didn't make a mistake.

Original equation: $ x^2 - 10x + y^2 + 6y + c = 0 $

Completing the square for x: $ x^2 -10x $ becomes $ (x - 5)^2 -25 $

For y: $ y^2 +6y $ becomes $ (y + 3)^2 -9 $

So substituting back:

$ (x - 5)^2 -25 + (y + 3)^2 -9 + c = 0 $

Combine constants: -25 -9 = -34, so:

$ (x - 5)^2 + (y + 3)^2 -34 + c = 0 $

Which rearranges to:

$ (x - 5)^2 + (y + 3)^2 = 34 - c $

Yes, that's correct. Then since radius squared is 1, so 34 - c = 1, so c = 34 -1 = 33. So c is 33.

Wait, but let me check if I didn't mix up any signs. Let me verify with another approach. Let me recall that the general equation $ x^2 + y^2 + Dx + Ey + F = 0 $ can be rewritten as $ (x + D/2)^2 + (y + E/2)^2 = (D/2)^2 + (E/2)^2 - F $. Therefore, the radius squared is $ (D/2)^2 + (E/2)^2 - F $.

In our case, the equation is $ x^2 -10x + y^2 +6y + c = 0 $. Comparing to the general form, D is -10, E is 6, and F is c. Therefore, radius squared is $ (-10/2)^2 + (6/2)^2 - c $, which is $ (-5)^2 + 3^2 - c = 25 + 9 - c = 34 - c $. Which matches what I had earlier. Therefore, radius squared is 34 - c. Since radius is 1, 34 - c = 1, so c = 33.

Therefore, the answer is 33. Let me check once more with plugging back in. Suppose c = 33, then the equation becomes:

$ x^2 -10x + y^2 +6y +33 = 0 $

Completing the square:

$ (x - 5)^2 -25 + (y + 3)^2 -9 +33 = 0 $

Which is $ (x - 5)^2 + (y + 3)^2 -25 -9 +33 = 0 $

Calculating constants: -25 -9 = -34, then -34 +33 = -1. Therefore:

$ (x - 5)^2 + (y + 3)^2 -1 = 0 $

Which is $ (x - 5)^2 + (y + 3)^2 = 1 $. So radius squared is 1, radius is 1. Perfect, that checks out.

Therefore, the value of c is indeed 33. I think that's correct. I don't see any mistakes in my reasoning. So I think the answer is 33.

**Final Answer**
The value of $ c $ is \boxed{33}.

To determine the value of $ c $ such that the circle defined by the equation

$$
x^2 - 10x + y^2 + 6y + c = 0
$$

has a radius of length 1, we begin by rewriting the equation in the standard form of a circle:

$$
(x - h)^2 + (y - k)^2 = r^2
$$

---

### Step 1: Complete the Square

We complete the square for both the $ x $ and $ y $ terms.

#### For the $ x $-terms:
$$
x^2 - 10x = (x - 5)^2 - 25
$$

#### For the $ y $-terms:
$$
y^2 + 6y = (y + 3)^2 - 9
$$

Substituting these into the original equation:

$$
(x - 5)^2 - 25 + (y + 3)^2 - 9 + c = 0
$$

Simplify the constants:

$$
(x - 5)^2 + (y + 3)^2 - 34 + c = 0
$$

Rearranging:

$$
(x - 5)^2 + (y + 3)^2 = 34 - c
$$

---

### Step 2: Use the Radius Condition

We are told the radius is 1, so the radius squared is:

$$
r^2 = 1^2 = 1
$$

From the equation above, the right-hand side is $ 34 - c $, so:

$$
34 - c = 1
$$

Solving for $ c $:

$$
c = 34 - 1 = 33
$$

---

### Final Answer

$$
\boxed{33}
$$

Solution: Completing the square gives us $(x - 5)^2 + (y + 3)^2 = 34 - c$. Since we want the radius to be 1, we must have $34 - c = 1^2$. It follows that $c = \boxed{33}$.


Explanation:
Okay, so I need to find the sum of the distances from the four intersection points of these two circles to the point (-3, 2). Let me start by understanding the problem step by step.

First, the equations given are:

1. $ x^2 + y^2 + 6x - 24y + 72 = 0 $
2. $ x^2 - y^2 + 6x + 16y - 46 = 0 $

They mention that these graphs intersect at four points. Wait, hold on. If they are both circles, then two circles can intersect at at most two points. So maybe there's a mistake here? Or perhaps one of them isn't a circle? Let me check.

Looking at the first equation: $ x^2 + y^2 + 6x - 24y + 72 = 0 $. This is a quadratic equation in x and y, and since the coefficients of $ x^2 $ and $ y^2 $ are both 1, it's a circle. Similarly, the second equation is $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Here, the coefficients of $ x^2 $ and $ y^2 $ are 1 and -1, respectively. So this is a hyperbola, not a circle. Therefore, the problem says they intersect at four points. Wait, but a circle and a hyperbola can intersect at up to four points. So that's possible. So maybe the first equation is a circle and the second is a hyperbola, and they intersect at four points. So the problem is correct.

So my task is to find the four points of intersection between the circle and the hyperbola, then compute the sum of the distances from each of these four points to the point (-3, 2). Then add them all up.

But solving for four points might be complicated. Maybe there's a smarter way than solving the system directly?

Alternatively, maybe I can find the equations of the curves, find their points of intersection, compute the distances, and sum them up. Let me try that approach.

First, let me rewrite both equations in standard form to understand their centers and radii (for the circle) and other properties (for the hyperbola).

Starting with the first equation: $ x^2 + y^2 + 6x - 24y + 72 = 0 $.

Let me complete the squares for x and y.

Group x terms and y terms:

$ (x^2 + 6x) + (y^2 - 24y) = -72 $

Complete the square for x:

Take coefficient of x, which is 6, divide by 2: 3, square it: 9. So add and subtract 9.

Similarly for y: coefficient of y is -24, divide by 2: -12, square it: 144. Add and subtract 144.

So:

$ (x^2 + 6x + 9 - 9) + (y^2 - 24y + 144 - 144) = -72 $

Which simplifies to:

$ (x + 3)^2 - 9 + (y - 12)^2 - 144 = -72 $

Combine constants:

$ (x + 3)^2 + (y - 12)^2 - 153 = -72 $

Add 153 to both sides:

$ (x + 3)^2 + (y - 12)^2 = 81 $

So the first equation is a circle with center at (-3, 12) and radius 9. Let me check that. The radius squared is 81, so radius is 9. Okay.

Now the second equation: $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Let me rearrange terms:

$ x^2 + 6x - y^2 + 16y = 46 $

Let me complete the square for x and y. Let me group x terms and y terms:

For x: $ x^2 + 6x $

For y: $ -y^2 + 16y $. Let me factor out the negative sign first: - (y^2 - 16y)

So:

$ (x^2 + 6x) - (y^2 - 16y) = 46 $

Complete the square for x:

Coefficient of x is 6, divide by 2: 3, square: 9. So add and subtract 9.

For y: inside the parentheses, y^2 - 16y. Take half of -16: -8, square: 64. So add and subtract 64.

So:

$ (x^2 + 6x + 9 - 9) - (y^2 - 16y + 64 - 64) = 46 $

Which becomes:

$ (x + 3)^2 - 9 - [(y - 8)^2 - 64] = 46 $

Distribute the negative sign:

$ (x + 3)^2 - 9 - (y - 8)^2 + 64 = 46 $

Combine constants:

$ (x + 3)^2 - (y - 8)^2 + 55 = 46 $

Subtract 55 from both sides:

$ (x + 3)^2 - (y - 8)^2 = -9 $

So the second equation is a hyperbola centered at (-3, 8), with the standard form $ \frac{(x + 3)^2}{a^2} - \frac{(y - 8)^2}{b^2} = 1 $, but here it's written as $ (x + 3)^2 - (y - 8)^2 = -9 $, which can be rewritten as $ \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 $. So it's a hyperbola centered at (-3, 8), opening upwards and downwards, with a^2 = 9 and b^2 = 9. So a = 3, b = 3. Interesting, so it's a rectangular hyperbola.

So now, the first equation is a circle centered at (-3, 12) with radius 9, and the second is a hyperbola centered at (-3, 8). The point (-3, 2) is given, which is along the vertical line x = -3, which is the line connecting the centers of the circle and hyperbola. The circle's center is at (-3, 12), and the hyperbola's center is at (-3, 8). The point (-3, 2) is below the hyperbola's center.

Now, the problem says that the graphs intersect at four points. Let me visualize this. The circle is centered at (-3, 12) with radius 9, so it goes from y = 12 - 9 = 3 to y = 12 + 9 = 21. The hyperbola is centered at (-3, 8), opening up and down. The circle's center is above the hyperbola's center. The hyperbola's equation is $ \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 $, so it has vertices at (-3, 8 ± 3), so (-3, 11) and (-3, 5). The circle's center is at (-3, 12), radius 9, so it reaches down to y = 3, which is below the hyperbola's lower vertex at y = 5. Therefore, the circle and hyperbola might intersect at four points. Let me check if that's possible.

But maybe I should proceed with solving the equations.

So, to find the points of intersection, I need to solve the system:

1. $ (x + 3)^2 + (y - 12)^2 = 81 $
2. $ (x + 3)^2 - (y - 8)^2 = -9 $

Let me denote equation 1 and equation 2. Let me subtract equation 2 from equation 1 to eliminate the squared terms. Wait, equation 1 is (x + 3)^2 + (y - 12)^2 = 81, and equation 2 is (x + 3)^2 - (y - 8)^2 = -9. If I subtract equation 2 from equation 1, I get:

[ (x + 3)^2 + (y - 12)^2 ] - [ (x + 3)^2 - (y - 8)^2 ] = 81 - (-9)

Simplify left side:

(x + 3)^2 cancels out. Then we have:

(y - 12)^2 - (- (y - 8)^2 ) = (y - 12)^2 + (y - 8)^2

Wait, no. Wait, subtracting equation 2 from equation 1:

Left side: [ (x + 3)^2 + (y - 12)^2 ] - [ (x + 3)^2 - (y - 8)^2 ] = (y - 12)^2 + (y - 8)^2

Right side: 81 - (-9) = 90

Therefore:

(y - 12)^2 + (y - 8)^2 = 90

Let me compute that:

First, expand (y - 12)^2 = y² - 24y + 144

(y - 8)^2 = y² - 16y + 64

Adding them together:

y² -24y +144 + y² -16y +64 = 2y² -40y + 208

Set equal to 90:

2y² -40y + 208 = 90

Subtract 90:

2y² -40y + 118 = 0

Divide both sides by 2:

y² -20y + 59 = 0

Now, solving for y:

Use quadratic formula: y = [20 ± sqrt(400 - 236)] / 2

Compute discriminant: 400 - 236 = 164

So sqrt(164) = sqrt(4*41) = 2*sqrt(41)

Therefore, y = [20 ± 2√41]/2 = 10 ± √41

So the y-coordinates of the intersection points are 10 + √41 and 10 - √41.

Now, let me find the corresponding x-values. To do that, I can use equation 2:

(x + 3)^2 - (y - 8)^2 = -9

Let me rearrange this equation:

(x + 3)^2 = (y - 8)^2 - 9

So once I have y, I can compute (y - 8)^2 - 9, then take square roots to find x + 3.

Alternatively, since we have two values of y, let me compute for each y.

First, let me compute for y = 10 + √41:

Compute (y - 8) = (10 + √41 - 8) = 2 + √41

Then (y - 8)^2 = (2 + √41)^2 = 4 + 4√41 + 41 = 45 + 4√41

Then (x + 3)^2 = (y - 8)^2 - 9 = 45 + 4√41 - 9 = 36 + 4√41

Therefore, x + 3 = ±√(36 + 4√41)

Similarly, for y = 10 - √41:

(y - 8) = 10 - √41 - 8 = 2 - √41

(y - 8)^2 = (2 - √41)^2 = 4 - 4√41 + 41 = 45 - 4√41

Then (x + 3)^2 = (y - 8)^2 - 9 = 45 - 4√41 - 9 = 36 - 4√41

Therefore, x + 3 = ±√(36 - 4√41)

So, for each y, there are two x-values. Therefore, total four points of intersection.

Therefore, the four points are:

For y = 10 + √41:

x = -3 ± √(36 + 4√41)

For y = 10 - √41:

x = -3 ± √(36 - 4√41)

So, these are the four points. Now, the problem asks for the sum of the distances from these four points to (-3, 2). Let me denote the point (-3, 2) as point P.

So, for each of the four points, compute the distance to P, then add them up.

But computing each distance individually might be tedious. Maybe there's a way to simplify this?

Let me recall that the distance from a point (x, y) to (-3, 2) is sqrt[(x + 3)^2 + (y - 2)^2]. Therefore, the sum of distances is sum_{i=1}^4 sqrt[(x_i + 3)^2 + (y_i - 2)^2].

But note that for each of the four points, (x + 3) is either sqrt(36 + 4√41) or -sqrt(36 + 4√41) or sqrt(36 - 4√41) or -sqrt(36 - 4√41). However, since we are squaring (x + 3)^2, maybe there's some symmetry here?

Alternatively, perhaps we can find expressions for (x + 3)^2 and (y - 2)^2 for each point?

Alternatively, maybe we can use the equations of the circle and hyperbola to find relations between (x + 3) and (y - 2)?

Alternatively, maybe we can use coordinates relative to the center of the circle?

Wait, the circle is centered at (-3, 12). Let me think. The point (-3, 2) is along the vertical line x = -3, which is the same line as the centers of both the circle and hyperbola. The distance from (-3, 2) to the center of the circle (-3, 12) is |12 - 2| = 10 units. The radius of the circle is 9, so the point (-3, 2) is outside the circle? Wait, because the distance from (-3, 2) to the center (-3, 12) is 10, which is greater than the radius 9. Therefore, (-3, 2) is outside the circle.

But how does this help?

Alternatively, maybe we can parametrize the points?

Alternatively, think about the four points lying on both the circle and hyperbola. Let me think about the distance from each point to (-3, 2). Let me denote that distance as D_i for each point.

But maybe there's a way to relate D_i with the equations?

Alternatively, since we have expressions for (x + 3)^2 and (y - 8)^2 from the hyperbola equation?

Wait, from equation 2: (x + 3)^2 - (y - 8)^2 = -9, so (x + 3)^2 = (y - 8)^2 - 9.

Also, from the circle equation: (x + 3)^2 + (y - 12)^2 = 81. So, substituting (x + 3)^2 from equation 2 into the circle equation:

[(y - 8)^2 - 9] + (y - 12)^2 = 81

Which is the same equation we had earlier, leading to y² -20y + 59 = 0. So that's consistent.

But how does this help with the distance to (-3, 2)?

Let me think. The distance squared from a point (x, y) to (-3, 2) is:

(x + 3)^2 + (y - 2)^2.

Let me denote this as D² = (x + 3)^2 + (y - 2)^2.

But from the hyperbola equation, we have (x + 3)^2 = (y - 8)^2 - 9. Therefore, substituting into D²:

D² = [(y - 8)^2 - 9] + (y - 2)^2

Let me compute that:

Expand (y - 8)^2 = y² - 16y + 64

(y - 2)^2 = y² - 4y + 4

Therefore:

D² = (y² - 16y + 64 - 9) + (y² - 4y + 4) = (y² -16y + 55) + (y² -4y + 4) = 2y² -20y + 59

Wait, that's interesting. So D² = 2y² -20y + 59. But earlier, when we solved for y, we had the equation y² -20y + 59 = 0. Wait, so if we have y² -20y + 59 = 0, then 2y² -20y + 59 = y² + (y² -20y + 59) = y² + 0 = y². Wait, that seems conflicting. Wait let me check:

Wait, D² = 2y² -20y + 59. But from the previous equation, we had y² -20y + 59 = 0. Therefore, substituting y² = 20y - 59 into D²:

D² = 2*(20y - 59) -20y + 59 = 40y - 118 -20y + 59 = 20y - 59.

Wait, so D² = 20y - 59?

Wait, that seems like a useful relation. Let me check again:

Original D² = 2y² -20y + 59.

But from the equation we had earlier, when we subtracted the two equations, we had y² -20y + 59 = 0. Therefore, y² = 20y - 59. Therefore, substituting into D²:

D² = 2*(20y - 59) -20y + 59 = 40y - 118 -20y + 59 = 20y - 59. Yes, that's correct.

Therefore, D² = 20y - 59. Therefore, D = sqrt(20y - 59). Therefore, the distance from each point to (-3, 2) is sqrt(20y - 59). Therefore, the sum of distances is sum_{i=1}^4 sqrt(20y_i - 59), where y_i are the y-coordinates of the four points, which are 10 + √41 and 10 - √41, each with two points.

But since for each y, there are two points (with ±x), but since the distance formula only depends on y, and the x-values are symmetric around -3, so for each y, the two points have the same distance to (-3, 2). Therefore, the sum of distances would be 2*sqrt(20*(10 + √41) - 59) + 2*sqrt(20*(10 - √41) - 59).

Let me compute that.

First, compute 20*(10 + √41) - 59:

= 200 + 20√41 - 59 = 141 + 20√41

Similarly, 20*(10 - √41) - 59 = 200 - 20√41 - 59 = 141 - 20√41

Therefore, the sum of distances is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41)

So, factor out the 2:

2[ sqrt(141 + 20√41) + sqrt(141 - 20√41) ]

Now, need to compute this expression. Let me denote A = sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me compute A.

Let me square A to simplify:

A² = [sqrt(141 + 20√41) + sqrt(141 - 20√41)]²

= (sqrt(141 + 20√41))² + 2*sqrt(141 + 20√41)*sqrt(141 - 20√41) + (sqrt(141 - 20√41))²

= (141 + 20√41) + 2*sqrt[(141 + 20√41)(141 - 20√41)] + (141 - 20√41)

Simplify:

= 141 + 20√41 + 141 - 20√41 + 2*sqrt[(141)^2 - (20√41)^2]

= 282 + 2*sqrt[141² - 400*41]

Compute the term inside the square root:

First compute 141²:

141² = (140 + 1)^2 = 140² + 2*140*1 + 1 = 19600 + 280 + 1 = 19881

Then compute 400*41 = 16400

Therefore, inside sqrt: 19881 - 16400 = 3481

Now sqrt(3481). Let me check what is sqrt(3481). Let me note that 59² = 3481 because 60² = 3600, so 59² = 3481. Yes, because 60² = 3600, subtract 2*60 +1 = 121, so 3600 - 121 = 3479? Wait, no, that formula is for (a - 1)^2 = a² - 2a + 1. Wait, 59² = (60 - 1)^2 = 60² - 2*60 + 1 = 3600 - 120 + 1 = 3481. Yes, correct.

Therefore, sqrt(3481) = 59.

Therefore, A² = 282 + 2*59 = 282 + 118 = 400

Therefore, A = sqrt(400) = 20. Since A is a sum of square roots, which are positive, so A = 20.

Therefore, the sum of distances is 2*A = 2*20 = 40.

Wait, so the total sum is 40? Let me check again.

Wait, the sum of distances was 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*A, and we found A = 20, so total sum is 40. Therefore, the answer is 40?

But let me verify this with another approach to make sure.

Alternatively, since we found that D² = 20y - 59, and the sum of distances is sum sqrt(20y_i - 59). But since for each y, there are two points, so sum sqrt(20y - 59) for each y, multiplied by 2. But we found that sum sqrt(20y - 59) over the four points is 2*A, which became 40. But let me check with actual numbers.

Alternatively, maybe there's a geometric interpretation. The point (-3, 2) is located along the line connecting the centers of the circle and hyperbola. The circle is centered at (-3, 12), radius 9. The distance from (-3, 2) to the center is 10, which is greater than the radius, so it's outside. The hyperbola is centered at (-3, 8). The point (-3, 2) is 6 units below the hyperbola's center.

But how does this relate? Maybe the four points lie on some relation with respect to (-3, 2). But perhaps not necessary.

Alternatively, since we found that the sum of distances is 40, which is a nice number, maybe that's correct. Let me check with specific values.

Let me compute sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me compute approximate numerical values to check if they sum to 20.

First, compute √41 ≈ 6.4031

Then 20√41 ≈ 20 * 6.4031 ≈ 128.062

Then 141 + 128.062 ≈ 269.062, sqrt(269.062) ≈ 16.403

Similarly, 141 - 128.062 ≈ 12.938, sqrt(12.938) ≈ 3.597

Adding them: 16.403 + 3.597 ≈ 20.0. Perfect! So that checks out. Therefore, the sum is indeed 20, multiplied by 2 gives 40. Therefore, the answer is 40.

Therefore, the sum of the distances is 40. Therefore, the answer is \boxed{40}.

But let me think again if there's another way to get this result, maybe using properties of circles and hyperbolas?

Alternatively, since we found that D² = 20y - 59, and we know the y-values of the intersection points satisfy y² -20y + 59 = 0. Therefore, for each y, y² = 20y - 59. Therefore, if we consider D² = 20y - 59, then D² = y². Therefore, D = |y|. Wait, but D is a distance, so positive. Therefore, D = |y|. Wait, but that would mean that the distance from each point to (-3, 2) is |y|. But that seems conflicting with previous calculations. Wait, let me check.

Wait, D² = 20y - 59, and y² = 20y - 59. Therefore, D² = y². Therefore, D = |y|. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum of distances is sum |y_i|. But since the y-coordinates are 10 + √41 and 10 - √41, each with two points. However, since √41 ≈ 6.4, 10 - √41 ≈ 3.6, which is positive. Therefore, all y_i are positive, so |y_i| = y_i. Therefore, the sum of distances is (10 + √41) + (10 + √41) + (10 - √41) + (10 - √41) = 4*10 + 2√41 - 2√41 = 40. So that's another way to see it! Therefore, the sum is 40. That's a much simpler way. So why did we go through all that?

Because we found that D² = y², so D = |y|. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum is sum |y_i|. Since all y_i are positive, sum y_i. Since there are two points with y = 10 + √41 and two points with y = 10 - √41, the total sum is 2*(10 + √41) + 2*(10 - √41) = 2*10 + 2*√41 + 2*10 - 2*√41 = 40. So that's straightforward. Therefore, the answer is 40. That's a much simpler approach. So why did we go through the previous steps? Because I didn't realize that D² = y². But how did that happen?

Earlier, we had D² = 20y - 59, and from the equation y² -20y + 59 = 0, so y² = 20y - 59. Therefore, D² = y². Therefore, D = |y|. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum is sum |y_i|. Since all y_i are positive, sum y_i. Therefore, the answer is 40. That's a much more straightforward way. So this shows that the answer is indeed 40.

But why is D² = y²? Let me check with coordinates. Let me take one of the points. Let me pick a point with y = 10 + √41. Let me compute the distance from (-3, 2) to this point. The distance squared is (x + 3)^2 + (y - 2)^2. From earlier, we had (x + 3)^2 = (y - 8)^2 - 9. Therefore, distance squared is [(y - 8)^2 - 9] + (y - 2)^2. Which we found equals y². Therefore, distance squared is y², so distance is |y|. Therefore, that's why. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum is sum |y_i|, which is sum y_i since all y_i are positive. Therefore, the answer is 40. That's a much better way to think about it.

Therefore, the answer is 40. Therefore, the sum of the distances is 40. Therefore, the answer is \boxed{40}.

**Final Answer**
\boxed{40}

To solve the problem, we are given two equations:

1. $ x^2 + y^2 + 6x - 24y + 72 = 0 $
2. $ x^2 - y^2 + 6x + 16y - 46 = 0 $

We are asked to compute the **sum of the distances** from the **four points of intersection** of these two curves to the point $(-3, 2)$.

---

### Step 1: Identify the Nature of the Curves

- The first equation is a **circle**. Completing the square gives:
  $$
  (x + 3)^2 + (y - 12)^2 = 81
  $$
  So, it's a circle centered at $(-3, 12)$ with radius 9.

- The second equation is a **hyperbola**. Completing the square gives:
  $$
  (x + 3)^2 - (y - 8)^2 = -9
  $$
  This is a hyperbola centered at $(-3, 8)$, opening vertically.

---

### Step 2: Find the Points of Intersection

We solve the system of equations:

1. $ (x + 3)^2 + (y - 12)^2 = 81 $
2. $ (x + 3)^2 - (y - 8)^2 = -9 $

Subtracting the second equation from the first:

$$
[(x + 3)^2 + (y - 12)^2] - [(x + 3)^2 - (y - 8)^2] = 81 - (-9)
$$
$$
(y - 12)^2 + (y - 8)^2 = 90
$$

Expanding both squares:

$$
(y^2 - 24y + 144) + (y^2 - 16y + 64) = 90
$$
$$
2y^2 - 40y + 208 = 90
$$
$$
2y^2 - 40y + 118 = 0
$$
$$
y^2 - 20y + 59 = 0
$$

Solving this quadratic:

$$
y = \frac{20 \pm \sqrt{400 - 236}}{2} = \frac{20 \pm \sqrt{164}}{2} = 10 \pm \sqrt{41}
$$

So the y-coordinates of the intersection points are $ y = 10 + \sqrt{41} $ and $ y = 10 - \sqrt{41} $, each with two corresponding x-values (due to the ± in the square root).

---

### Step 3: Compute the Sum of Distances

We are to compute the **sum of the distances** from each of the four points to the point $(-3, 2)$.

Let’s denote a general point of intersection as $(x, y)$. The distance from this point to $(-3, 2)$ is:

$$
D = \sqrt{(x + 3)^2 + (y - 2)^2}
$$

From the second equation, we can express $(x + 3)^2$ in terms of $y$:

$$
(x + 3)^2 = (y - 8)^2 - 9
$$

Substitute into the distance formula:

$$
D^2 = (x + 3)^2 + (y - 2)^2 = [(y - 8)^2 - 9] + (y - 2)^2
$$

Now expand both terms:

- $(y - 8)^2 = y^2 - 16y + 64$
- $(y - 2)^2 = y^2 - 4y + 4$

So:

$$
D^2 = (y^2 - 16y + 64 - 9) + (y^2 - 4y + 4) = 2y^2 - 20y + 59
$$

But from the earlier quadratic equation, we know:

$$
y^2 - 20y + 59 = 0 \Rightarrow y^2 = 20y - 59
$$

Substitute into $D^2$:

$$
D^2 = 2y^2 - 20y + 59 = 2(20y - 59) - 20y + 59 = 40y - 118 - 20y + 59 = 20y - 59
$$

But we also know that $y^2 = 20y - 59$, so:

$$
D^2 = y^2 \Rightarrow D = |y|
$$

Thus, the **distance from each point to $(-3, 2)$ is simply $|y|$**.

---

### Step 4: Sum the Distances

We have two distinct y-values: $10 + \sqrt{41}$ and $10 - \sqrt{41}$, each occurring twice (since for each y, there are two x-values). So the total sum of distances is:

$$
2 \cdot (10 + \sqrt{41}) + 2 \cdot (10 - \sqrt{41}) = 20 + 2\sqrt{41} + 20 - 2\sqrt{41} = 40
$$

---

### Final Answer

$$
\boxed{40}
$$

[Assistant]
Explanation:
Okay, so I need to find the sum of the distances from the four intersection points of these two circles to the point (-3, 2). Let me start by understanding the problem step by step.

First, the equations given are:

1. $ x^2 + y^2 + 6x - 24y + 72 = 0 $
2. $ x^2 - y^2 + 6x + 16y - 46 = 0 $

They mention that these graphs intersect at four points. Wait, hold on. If they are both circles, then two circles can intersect at at most two points. So maybe there's a mistake here? Or perhaps one of them isn't a circle? Let me check.

Looking at the first equation: $ x^2 + y^2 + 6x - 24y + 72 = 0 $. This is a quadratic equation in x and y, and since the coefficients of $ x^2 $ and $ y^2 $ are both 1, it's a circle. Similarly, the second equation is $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Here, the coefficients of $ x^2 $ and $ y^2 $ are 1 and -1, respectively. So this is a hyperbola, not a circle. Therefore, the problem says they intersect at four points. Wait, but a circle and a hyperbola can intersect at up to four points. So that's possible. So maybe the first equation is a circle and the second is a hyperbola, and they intersect at four points. So the problem is correct.

So my task is to find the four points of intersection between the circle and the hyperbola, then compute the sum of the distances from each of these four points to the point (-3, 2). Then add them all up.

But solving for four points might be complicated. Maybe there's a smarter way than solving the system directly?

Alternatively, maybe I can find the equations of the curves, find their points of intersection, compute the distances, and sum them up. Let me try that approach.

First, let me rewrite both equations in standard form to understand their centers and radii (for the circle) and other properties (for the hyperbola).

Starting with the first equation: $ x^2 + y^2 + 6x - 24y + 72 = 0 $.

Let me complete the squares for x and y.

Group x terms and y terms:

$ (x^2 + 6x) + (y^2 - 24y) = -72 $

Complete the square for x:

Take coefficient of x, which is 6, divide by 2: 3, square it: 9. So add and subtract 9.

Similarly for y: coefficient of y is -24, divide by 2: -12, square it: 144. Add and subtract 144.

So:

$ (x^2 + 6x + 9 - 9) + (y^2 - 24y + 144 - 144) = -72 $

Which simplifies to:

$ (x + 3)^2 - 9 + (y - 12)^2 - 144 = -72 $

Combine constants:

$ (x + 3)^2 + (y - 12)^2 - 153 = -72 $

Add 153 to both sides:

$ (x + 3)^2 + (y - 12)^2 = 81 $

So the first equation is a circle with center at (-3, 12) and radius 9. Let me check that. The radius squared is 81, so radius is 9. Okay.

Now the second equation: $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Let me rearrange terms:

$ x^2 + 6x - y^2 + 16y = 46 $

Let me complete the square for x and y. Let me group x terms and y terms:

For x: $ x^2 + 6x $

For y: $ -y^2 + 16y $. Let me factor out the negative sign first: - (y^2 - 16y)

So:

$ (x^2 + 6x) - (y^2 - 16y) = 46 $

Complete the square for x:

Coefficient of x is 6, divide by 2: 3, square: 9. So add and subtract 9.

For y: inside the parentheses, y^2 - 16y. Take half of -16: -8, square: 64. So add and subtract 64.

So:

$ (x^2 + 6x + 9 - 9) - (y^2 - 16y + 64 - 64) = 46 $

Which becomes:

$ (x + 3)^2 - 9 - [(y - 8)^2 - 64] = 46 $

Distribute the negative sign:

$ (x + 3)^2 - 9 - (y - 8)^2 + 64 = 46 $

Combine constants:

$ (x + 3)^2 - (y - 8)^2 + 55 = 46 $

Subtract 55 from both sides:

$ (x + 3)^2 - (y - 8)^2 = -9 $

So the second equation is a hyperbola centered at (-3, 8), with the standard form $ \frac{(x + 3)^2}{a^2} - \frac{(y - 8)^2}{b^2} = 1 $, but here it's written as $ (x + 3)^2 - (y - 8)^2 = -9 $, which can be rewritten as $ \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 $. So it's a hyperbola centered at (-3, 8), opening upwards and downwards, with a^2 = 9 and b^2 = 9. So a = 3, b = 3. Interesting, so it's a rectangular hyperbola.

So now, the first equation is a circle centered at (-3, 12) with radius 9, and the second is a hyperbola centered at (-3, 8). The point (-3, 2) is given, which is along the vertical line x = -3, which is the line connecting the centers of the circle and hyperbola. The circle's center is at (-3, 12), and the hyperbola's center is at (-3, 8). The point (-3, 2) is below the hyperbola's center.

Now, the problem says that the graphs intersect at four points. Let me visualize this. The circle is centered at (-3, 12) with radius 9, so it goes from y = 12 - 9 = 3 to y = 12 + 9 = 21. The hyperbola is centered at (-3, 8), opening up and down. The circle's center is above the hyperbola's center. The hyperbola's equation is $ \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 $, so it has vertices at (-3, 8 ± 3), so (-3, 11) and (-3, 5). The circle's center is at (-3, 12), radius 9, so it reaches down to y = 3, which is below the hyperbola's lower vertex at y = 5. Therefore, the circle and hyperbola might intersect at four points. Let me check if that's possible.

But maybe I should proceed with solving the equations.

So, to find the points of intersection, I need to solve the system:

1. $ (x + 3)^2 + (y - 12)^2 = 81 $
2. $ (x + 3)^2 - (y - 8)^2 = -9 $

Let me denote equation 1 and equation 2. Let me subtract equation 2 from equation 1 to eliminate the squared terms. Wait, equation 1 is (x + 3)^2 + (y - 12)^2 = 81, and equation 2 is (x + 3)^2 - (y - 8)^2 = -9. If I subtract equation 2 from equation 1, I get:

[ (x + 3)^2 + (y - 12)^2 ] - [ (x + 3)^2 - (y - 8)^2 ] = 81 - (-9)

Simplify left side:

(x + 3)^2 cancels out. Then we have:

(y - 12)^2 - (- (y - 8)^2 ) = (y - 12)^2 + (y - 8)^2

Wait, no. Wait, subtracting equation 2 from equation 1:

Left side: [ (x + 3)^2 + (y - 12)^2 ] - [ (x + 3)^2 - (y - 8)^2 ] = (y - 12)^2 + (y - 8)^2

Right side: 81 - (-9) = 90

Therefore:

(y - 12)^2 + (y - 8)^2 = 90

Let me compute that:

First, expand (y - 12)^2 = y² - 24y + 144

(y - 8)^2 = y² - 16y + 64

Adding them together:

y² -24y +144 + y² -16y +64 = 2y² -40y + 208

Set equal to 90:

2y² -40y + 208 = 90

Subtract 90:

2y² -40y + 118 = 0

Divide both sides by 2:

y² -20y + 59 = 0

Now, solving for y:

Use quadratic formula: y = [20 ± sqrt(400 - 236)] / 2

Compute discriminant: 400 - 236 = 164

So sqrt(164) = sqrt(4*41) = 2*sqrt(41)

Therefore, y = [20 ± 2√41]/2 = 10 ± √41

So the y-coordinates of the intersection points are 10 + √41 and 10 - √41.

Now, let me find the corresponding x-values. To do that, I can use equation 2:

(x + 3)^2 - (y - 8)^2 = -9

Let me rearrange this equation:

(x + 3)^2 = (y - 8)^2 - 9

So once I have y, I can compute (y - 8)^2 - 9, then take square roots to find x + 3.

Alternatively, since we have two values of y, let me compute for each y.

First, let me compute for y = 10 + √41:

Compute (y - 8) = (10 + √41 - 8) = 2 + √41

Then (y - 8)^2 = (2 + √41)^2 = 4 + 4√41 + 41 = 45 + 4√41

Then (x + 3)^2 = (y - 8)^2 - 9 = 45 + 4√41 - 9 = 36 + 4√41

Therefore, x + 3 = ±√(36 + 4√41)

Similarly, for y = 10 - √41:

(y - 8) = 10 - √41 - 8 = 2 - √41

(y - 8)^2 = (2 - √41)^2 = 4 - 4√41 + 41 = 45 - 4√41

Then (x + 3)^2 = (y - 8)^2 - 9 = 45 - 4√41 - 9 = 36 - 4√41

Therefore, x + 3 = ±√(36 - 4√41)

So, for each y, there are two x-values. Therefore, total four points of intersection.

Therefore, the four points are:

For y = 10 + √41:

x = -3 ± √(36 + 4√41)

For y = 10 - √41:

x = -3 ± √(36 - 4√41)

So, these are the four points. Now, the problem asks for the sum of the distances from these four points to (-3, 2). Let me denote the point (-3, 2) as point P.

So, for each of the four points, compute the distance to P, then add them up.

But computing each distance individually might be tedious. Maybe there's a way to simplify this?

Let me recall that the distance from a point (x, y) to (-3, 2) is sqrt[(x + 3)^2 + (y - 2)^2]. Therefore, the sum of distances is sum_{i=1}^4 sqrt[(x_i + 3)^2 + (y_i - 2)^2].

But note that for each of the four points, (x + 3) is either sqrt(36 + 4√41) or -sqrt(36 + 4√41) or sqrt(36 - 4√41) or -sqrt(36 - 4√41). However, since we are squaring (x + 3)^2, maybe there's some symmetry here?

Alternatively, perhaps we can find expressions for (x + 3)^2 and (y - 2)^2 for each point?

Alternatively, maybe we can use the equations of the circle and hyperbola to find relations between (x + 3) and (y - 2)?

Alternatively, maybe we can use coordinates relative to the center of the circle?

Wait, the circle is centered at (-3, 12). Let me think. The point (-3, 2) is along the vertical line x = -3, which is the same line as the centers of both the circle and hyperbola. The distance from (-3, 2) to the center of the circle (-3, 12) is |12 - 2| = 10 units. The radius of the circle is 9, so the point (-3, 2) is outside the circle? Wait, because the distance from (-3, 2) to the center (-3, 12) is 10, which is greater than the radius 9. Therefore, (-3, 2) is outside the circle.

But how does this help?

Alternatively, maybe we can parametrize the points?

Alternatively, think about the four points lying on both the circle and hyperbola. Let me think about the distance from each point to (-3, 2). Let me denote that distance as D_i for each point.

But maybe there's a way to relate D_i with the equations?

Alternatively, since we have expressions for (x + 3)^2 and (y - 8)^2 from the hyperbola equation?

Wait, from equation 2: (x + 3)^2 - (y - 8)^2 = -9, so (x + 3)^2 = (y - 8)^2 - 9.

Also, from the circle equation: (x + 3)^2 + (y - 12)^2 = 81. So, substituting (x + 3)^2 from equation 2 into the circle equation:

[(y - 8)^2 - 9] + (y - 12)^2 = 81

Which is the same equation we had earlier, leading to y² -20y + 59 = 0. So that's consistent.

But how does this help with the distance to (-3, 2)?

Let me think. The distance squared from a point (x, y) to (-3, 2) is:

(x + 3)^2 + (y - 2)^2.

Let me denote this as D² = (x + 3)^2 + (y - 2)^2.

But from the hyperbola equation, we have (x + 3)^2 = (y - 8)^2 - 9. Therefore, substituting into D²:

D² = [(y - 8)^2 - 9] + (y - 2)^2

Let me compute that:

Expand (y - 8)^2 = y² - 16y + 64

(y - 2)^2 = y² - 4y + 4

Therefore:

D² = (y² - 16y + 64 - 9) + (y² - 4y + 4) = (y² -16y + 55) + (y² -4y + 4) = 2y² -20y + 59

Wait, that's interesting. So D² = 2y² -20y + 59. But earlier, when we solved for y, we had the equation y² -20y + 59 = 0. Wait, so if we have y² -20y + 59 = 0, then 2y² -20y + 59 = y² + (y² -20y + 59) = y² + 0 = y². Wait, that seems conflicting. Wait let me check:

Wait, D² = 2y² -20y + 59. But from the previous equation, we had y² -20y + 59 = 0. Therefore, substituting y² = 20y - 59 into D²:

D² = 2*(20y - 59) -20y + 59 = 40y - 118 -20y + 59 = 20y - 59.

Wait, so D² = 20y - 59?

Wait, that seems like a useful relation. Let me check again:

Original D² = 2y² -20y + 59.

But from the equation we had earlier, when we subtracted the two equations, we had y² -20y + 59 = 0. Therefore, y² = 20y - 59. Therefore, substituting into D²:

D² = 2*(20y - 59) -20y + 59 = 40y - 118 -20y + 59 = 20y - 59. Yes, that's correct.

Therefore, D² = 20y - 59. Therefore, D = sqrt(20y - 59). Therefore, the distance from each point to (-3, 2) is sqrt(20y - 59). Therefore, the sum of distances is sum_{i=1}^4 sqrt(20y_i - 59), where y_i are the y-coordinates of the four points, which are 10 + √41 and 10 - √41, each with two points.

But since for each y, there are two points (with ±x), but since the distance formula only depends on y, and the x-values are symmetric around -3, so for each y, the two points have the same distance to (-3, 2). Therefore, the sum of distances would be 2*sqrt(20*(10 + √41) - 59) + 2*sqrt(20*(10 - √41) - 59).

Let me compute that.

First, compute 20*(10 + √41) - 59:

= 200 + 20√41 - 59 = 141 + 20√41

Similarly, 20*(10 - √41) - 59 = 200 - 20√41 - 59 = 141 - 20√41

Therefore, the sum of distances is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41)

So, factor out the 2:

2[ sqrt(141 + 20√41) + sqrt(141 - 20√41) ]

Now, need to compute this expression. Let me denote A = sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me compute A.

Let me square A to simplify:

A² = [sqrt(141 + 20√41) + sqrt(141 - 20√41)]²

= (sqrt(141 + 20√41))² + 2*sqrt(141 + 20√41)*sqrt(141 - 20√41) + (sqrt(141 - 20√41))²

= (141 + 20√41) + 2*sqrt[(141 + 20√41)(141 - 20√41)] + (141 - 20√41)

Simplify:

= 141 + 20√41 + 141 - 20√41 + 2*sqrt[(141)^2 - (20√41)^2]

= 282 + 2*sqrt[141² - 400*41]

Compute the term inside the square root:

First compute 141²:

141² = (140 + 1)^2 = 140² + 2*140*1 + 1 = 19600 + 280 + 1 = 19881

Then compute 400*41 = 16400

Therefore, inside sqrt: 19881 - 16400 = 3481

Now sqrt(3481). Let me check what is sqrt(3481). Let me note that 59² = 3481 because 60² = 3600, so 59² = 3481. Yes, because 60² = 3600, subtract 2*60 +1 = 121, so 3600 - 121 = 3479? Wait, no, that formula is for (a - 1)^2 = a² - 2a + 1. Wait, 59² = (60 - 1)^2 = 60² - 2*60 + 1 = 3600 - 120 + 1 = 3481. Yes, correct.

Therefore, sqrt(3481) = 59.

Therefore, A² = 282 + 2*59 = 282 + 118 = 400

Therefore, A = sqrt(400) = 20. Since A is a sum of square roots, which are positive, so A = 20.

Therefore, the sum of distances is 2*A = 2*20 = 40.

Wait, so the total sum is 40? Let me check again.

Wait, the sum of distances was 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*A, and we found A = 20, so total sum is 40. Therefore, the answer is 40?

But let me verify this with another approach to make sure.

Alternatively, since we found that D² = 20y - 59, and the sum of distances is sum sqrt(20y_i - 59). But since for each y, there are two points, so sum sqrt(20y - 59) for each y, multiplied by 2. But we found that sum sqrt(20y - 59) over the four points is 2*A, which became 40. But let me check with actual numbers.

Alternatively, maybe there's a geometric interpretation. The point (-3, 2) is located along the line connecting the centers of the circle and hyperbola. The circle is centered at (-3, 12), radius 9. The distance from (-3, 2) to the center is 10, which is greater than the radius, so it's outside. The hyperbola is centered at (-3, 8). The point (-3, 2) is 6 units below the hyperbola's center.

But how does this relate? Maybe the four points lie on some relation with respect to (-3, 2). But perhaps not necessary.

Alternatively, since we found that the sum of distances is 40, which is a nice number, maybe that's correct. Let me check with specific values.

Let me compute sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me compute approximate numerical values to check if they sum to 20.

First, compute √41 ≈ 6.4031

Then 20√41 ≈ 20 * 6.4031 ≈ 128.062

Then 141 + 128.062 ≈ 269.062, sqrt(269.062) ≈ 16.403

Similarly, 141 - 128.062 ≈ 12.938, sqrt(12.938) ≈ 3.597

Adding them: 16.403 + 3.597 ≈ 20.0. Perfect! So that checks out. Therefore, the sum is indeed 20, multiplied by 2 gives 40. Therefore, the answer is 40.

Therefore, the sum of the distances is 40. Therefore, the answer is \boxed{40}.

But let me think again if there's another way to get this result, maybe using properties of circles and hyperbolas?

Alternatively, since we found that D² = 20y - 59, and we know the y-values of the intersection points satisfy y² -20y + 59 = 0. Therefore, for each y, y² = 20y - 59. Therefore, if we consider D² = 20y - 59, then D² = y². Therefore, D = |y|. Wait, but D is a distance, so positive. Therefore, D = |y|. Wait, but that would mean that the distance from each point to (-3, 2) is |y|. But that seems conflicting with previous calculations. Wait, let me check.

Wait, D² = 20y - 59, and y² = 20y - 59. Therefore, D² = y². Therefore, D = |y|. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum of distances is sum |y_i|. But since the y-coordinates are 10 + √41 and 10 - √41, each with two points. However, since √41 ≈ 6.4, 10 - √41 ≈ 3.6, which is positive. Therefore, all y_i are positive, so |y_i| = y_i. Therefore, the sum of distances is (10 + √41) + (10 + √41) + (10 - √41) + (10 - √41) = 4*10 + 2√41 - 2√41 = 40. So that's another way to see it! Therefore, the sum is 40. That's a much simpler way. So why did we go through all that?

Because we found that D² = y², so D = |y|. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum is sum |y_i|. Since all y_i are positive, sum y_i. Since there are two points with y = 10 + √41 and two points with y = 10 - √41, the total sum is 2*(10 + √41) + 2*(10 - √41) = 2*10 + 2*√41 + 2*10 - 2*√41 = 40. So that's straightforward. Therefore, the answer is 40. That's a much simpler approach. So why did we go through the previous steps? Because I didn't realize that D² = y². But how did that happen?

Earlier, we had D² = 20y - 59, and from the equation y² -20y + 59 = 0, so y² = 20y - 59. Therefore, D² = y². Therefore, D = |y|. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum is sum |y_i|. Since all y_i are positive, sum y_i. Therefore, the answer is 40. That's a much more straightforward way. So this shows that the answer is indeed 40.

But why is D² = y²? Let me check with coordinates. Let me take one of the points. Let me pick a point with y = 10 + √41. Let me compute the distance from (-3, 2) to this point. The distance squared is (x + 3)^2 + (y - 2)^2. From earlier, we had (x + 3)^2 = (y - 8)^2 - 9. Therefore, distance squared is [(y - 8)^2 - 9] + (y - 2)^2. Which we found equals y². Therefore, distance squared is y², so distance is |y|. Therefore, that's why. Therefore, the distance from each point to (-3, 2) is |y|. Therefore, the sum is sum |y_i|, which is sum y_i since all y_i are positive. Therefore, the answer is 40. That's a much better way to think about it.

Therefore, the answer is 40. Therefore, the sum of the distances is 40. Therefore, the answer is \boxed{40}.

**Final Answer**
\boxed{40}

To solve the problem, we are given two equations:

1. $ x^2 + y^2 + 6x - 24y + 72 = 0 $
2. $ x^2 - y^2 + 6x + 16y - 46 = 0 $

We are asked to compute the **sum of the distances** from the **four points of intersection** of these two curves to the point $(-3, 2)$.

---

### Step 1: Identify the Nature of the Curves

- The first equation is a **circle**. Completing the square gives:
  $$
  (x + 3)^2 + (y - 12)^2 = 81
  $$
  So, it's a circle centered at $(-3, 12)$ with radius 9.

- The second equation is a **hyperbola**. Completing the square gives:
  $$
  (x + 3)^2 - (y - 8)^2 = -9
  $$
  This is a hyperbola centered at $(-3, 8)$, opening vertically.

---

### Step 2: Find the Points of Intersection

We solve the system of equations:

1. $ (x + 3)^2 + (y - 12)^2 = 81 $
2. $ (x + 3)^2 - (y - 8)^2 = -9 $

Subtracting the second equation from the first:

$$
[(x + 3)^2 + (y - 12)^2] - [(x + 3)^2 - (y - 8)^2] = 81 - (-9)
$$
$$
(y - 12)^2 + (y - 8)^2 = 90
$$

Expanding both squares:

$$
(y^2 - 24y + 144) + (y^2 - 16y + 64) = 90
$$
$$
2y^2 - 40y + 208 = 90
$$
$$
2y^2 - 40y + 118 = 0
$$
$$
y^2 - 20y + 59 = 0
$$

Solving this quadratic:

$$
y = \frac{20 \pm \sqrt{400 - 236}}{2} = \frac{20 \pm \sqrt{164}}{2} = 10 \pm \sqrt{41}
$$

So the y-coordinates of the intersection points are $ y = 10 + \sqrt{41} $ and $ y = 10 - \sqrt{41} $, each with two corresponding x-values (due to the ± in the square root).

---

### Step 3: Compute the Sum of Distances

We are to compute the **sum of the distances** from each of the four points to the point $(-3, 2)$.

Let’s denote a general point of intersection as $(x, y)$. The distance from this point to $(-3, 2)$ is:

$$
D = \sqrt{(x + 3)^2 + (y - 2)^2}
$$

From the second equation, we can express $(x + 3)^2$ in terms of $y$:

$$
(x + 3)^2 = (y - 8)^2 - 9
$$

Substitute into the distance formula:

$$
D^2 = (x + 3)^2 + (y - 2)^2 = [(y - 8)^2 - 9] + (y - 2)^2
$$

Now expand both terms:

- $(y - 8)^2 = y^2 - 16y + 64$
- $(y - 2)^2 = y^2 - 4y + 4$

So:

$$
D^2 = (y^2 - 16y + 64 - 9) + (y^2 - 4y + 4) = 2y^2 - 20y + 59
$$

But from the earlier quadratic equation, we know:

$$
y^2 - 20y + 59 = 0 \Rightarrow y^2 = 20y - 59
$$

Substitute into $D^2$:

$$
D^2 = 2y^2 - 20y + 59 = 2(20y - 59) - 20y + 59 = 40y - 118 - 20y + 59 = 20y - 59
$$

But we also know that $y^2 = 20y - 59$, so:

$$
D^2 = y^2 \Rightarrow D = |y|
$$

Thus, the **distance from each point to $(-3, 2)$ is simply $|y|$**.

---

### Step 4: Sum the Distances

We have two distinct y-values: $10 + \sqrt{41}$ and $10 - \sqrt{41}$, each occurring twice (since for each y, there are two x-values). So the total sum of distances is:

$$
2 \cdot (10 + \sqrt{41}) + 2 \cdot (10 - \sqrt{41}) = 20 + 2\sqrt{41} + 20 - 2\sqrt{41} = 40
$$

---

### Final Answer

$$
\boxed{40}
$$

Solution:
Adding the equations, we get
\[2x^2 + 12x - 8y + 26 = 0,\]or $x^2 + 6x - 4y + 13 = 0.$  We can write this equation as
\[(x + 3)^2 = 4(y - 1).\]This is the equation of the parabola with focus $(-3,2)$ and directrix $y = 0.$

[asy]
unitsize(1 cm);

real parab (real x) {
  return ((x^2 + 6*x + 13)/4);
}

pair P = (-0.5,parab(-0.5));

draw(graph(parab,-6,0));
draw((-3,2)--P--(-0.5,0));

dot((-3,2));
dot((-3,1));
draw((-6,0)--(0,0),dashed);
[/asy]

By definition of a parabola, for any point $P$ on the parabola, the distance from $P$ to the focus is equal to the distance from $P$ to the $y$-axis, which is the $y$-coordinate of the point.

Subtracting the given equations, we get $2y^2 - 40y + 118 = 0,$ or $y^2 - 20y + 59 = 0.$  Let $y_1$ and $y_2$ be the roots of this quadratic.  Then the $y$-coordinate of each point of intersection must be either $y_1$ or $y_2.$

Note that the equation $x^2 + y^2 + 6x - 24xy + 72 = 0$ represents a circle, so it intersects the line $y = y_1$ in at most two points, and the line $y = y_2$ is at most two points.  Therefore, the $y$-coordinates of the four points of intersection must be $y_1,$ $y_1,$ $y_2,$ $y_2,$ and their sum is $2y_1 + 2y_2.$

By Vieta's formulas, $y_1 + y_2 = 20,$ so $2y_1 + 2y_2 = \boxed{40}.$