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 the intersection points might be complicated. Let me think if there's a smarter way than solving the equations directly. Maybe using some geometric properties?

Alternatively, maybe I can subtract the equations to eliminate some terms. Let me try that.

First, let me write both equations again:

Equation 1: $ x^2 + y^2 + 6x - 24y + 72 = 0 $

Equation 2: $ x^2 - y^2 + 6x + 16y - 46 = 0 $

If I subtract Equation 2 from Equation 1, I can eliminate the $ x^2 $ and $ 6x $ terms. Let me do that:

Equation 1 - Equation 2:

$ (x^2 + y^2 + 6x - 24y + 72) - (x^2 - y^2 + 6x + 16y - 46) = 0 - 0 $

Simplify term by term:

x² - x² = 0

y² - (-y²) = y² + y² = 2y²

6x - 6x = 0

-24y - 16y = -40y

72 - (-46) = 72 + 46 = 118

So altogether:

2y² - 40y + 118 = 0

Divide both sides by 2:

y² - 20y + 59 = 0

Wait, let me check that again. Wait, 72 + 46 is 118? 72 + 46 is 118? 70 + 40 is 110, 2 + 6 is 8, so 118. Yes. Then dividing by 2 gives y² - 20y + 59 = 0. Let me check the coefficients again.

Original subtraction:

Equation 1: + y², Equation 2: - y², so Equation 1 - Equation 2: + y² - (-y²) = 2y². Then -24y - 16y is -40y. Then +72 - (-46) is +72 + 46 = 118. So yes, 2y² - 40y + 118 = 0. Dividing by 2: y² - 20y + 59 = 0. Let me check discriminant: D = 400 - 4*1*59 = 400 - 236 = 164. Which is positive, so two real solutions. Therefore, there are two y-values where the curves intersect. But the problem says four points. So maybe each y corresponds to two x-values? But since one equation is a circle and the other is a hyperbola, maybe each y gives two x's? Let me check.

Alternatively, maybe I need to solve for y first. Let me solve the quadratic equation for y:

y² - 20y + 59 = 0

Using quadratic formula:

y = [20 ± sqrt(400 - 236)] / 2 = [20 ± sqrt(164)] / 2

Simplify sqrt(164): sqrt(4*41) = 2*sqrt(41). Therefore:

y = [20 ± 2√41]/2 = 10 ± √41

So two y-values: 10 + √41 and 10 - √41. So each of these y-values will correspond to some x-values. Let me find x-values for each y.

But to find x, I need to substitute back into one of the original equations. Let me pick Equation 1 or Equation 2? Maybe Equation 1 is a circle, so maybe easier to write in standard form?

Alternatively, maybe I can use Equation 2? Let me see. Let me try Equation 1 first.

Equation 1: $ x^2 + y^2 + 6x - 24y + 72 = 0 $

Let me rearrange this equation to standard form by completing the squares.

Group x terms and y terms:

x² + 6x + y² - 24y = -72

Complete the square for x:

x² + 6x = (x + 3)^2 - 9

For y:

y² - 24y = (y - 12)^2 - 144

So substituting back:

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

Combine constants:

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

Therefore:

(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 is $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Let me try to write this in standard form. Since it's a hyperbola, maybe I can rearrange terms.

Group x terms and y terms:

x² + 6x - y² + 16y = 46

Complete the square for x and y:

For x: x² + 6x = (x + 3)^2 - 9

For y: -y² + 16y = - (y² - 16y) = -[(y - 8)^2 - 64] = - (y - 8)^2 + 64

So substituting back:

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

Combine constants:

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

Therefore:

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

Which can be written as:

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

So this is a hyperbola centered at (-3, 8), with transverse axis along the y-axis, since the y-term is positive. The standard form is [(y - k)^2]/a² - [(x - h)^2]/b² = 1. Here, [(y - 8)^2]/9 - [(x + 3)^2]/9 = 1. Wait, because 9 is the right-hand side. So a² = 9, so a = 3, and b² = 9, so b = 3. Therefore, the hyperbola has asymptotes y - 8 = ±(b/a)(x + 3) => y - 8 = ±(x + 3). So asymptotes are y = x + 11 and y = -x + 5.

But maybe I don't need that. Anyway, the key point is that the hyperbola is centered at (-3, 8), and the circle is centered at (-3, 12). So both centers have the same x-coordinate, -3. So they are vertically aligned. The circle has radius 9, and the hyperbola is centered at (-3, 8). Let me check the distance between centers. The distance between (-3, 12) and (-3, 8) is |12 - 8| = 4 units. Since the circle has radius 9, which is larger than 4, so the hyperbola is inside the circle? Wait, but hyperbola extends infinitely. However, the intersection points are four points. So maybe the hyperbola intersects the circle at four points.

But going back, I need to find the points of intersection. Earlier, I found that the y-values are 10 ± √41. Let me compute approximate values to check if they make sense. √41 is approx 6.4, so 10 + 6.4 = 16.4, and 10 - 6.4 = 3.6. So y-values are around 16.4 and 3.6. The circle is centered at (-3, 12) with radius 9, so the topmost point of the circle is at y = 12 + 9 = 21, and bottommost at y = 12 - 9 = 3. So y = 16.4 is within the circle's range (since 16.4 < 21), and y = 3.6 is also within the circle's range (since 3.6 > 3). So that seems okay.

Now, to find x-values for each y. Let me use Equation 1 or Equation 2? Maybe Equation 2? Let me check. Let me use Equation 2: $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Let me rearrange it as:

x² + 6x = y² - 16y + 46

But maybe it's better to use Equation 1. Let me try Equation 1 again. Since we have the circle equation in standard form, maybe we can use that. Let me recall that the circle equation is:

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

So if I know y, I can solve for x. Let me do that. Let me take the y-values we found: y = 10 ± √41.

Let me denote y = 10 + √41 first. Then plug into the circle equation:

(x + 3)^2 + ( (10 + √41) - 12 )^2 = 81

Simplify the y-term:

(10 + √41 - 12) = (-2 + √41)

Therefore:

(x + 3)^2 + (-2 + √41)^2 = 81

Compute (-2 + √41)^2:

= (-2)^2 + 2*(-2)*(√41) + (√41)^2

= 4 - 4√41 + 41

= 45 - 4√41

Therefore:

(x + 3)^2 + 45 - 4√41 = 81

So (x + 3)^2 = 81 - 45 + 4√41 = 36 + 4√41

Therefore, x + 3 = ±√(36 + 4√41)

Therefore, x = -3 ± √(36 + 4√41)

Similarly, for y = 10 - √41:

(x + 3)^2 + ( (10 - √41) - 12 )^2 = 81

Simplify the y-term:

(10 - √41 - 12) = (-2 - √41)

Therefore:

(x + 3)^2 + (-2 - √41)^2 = 81

Compute (-2 - √41)^2:

= (-2)^2 + 2*(-2)*(-√41) + (√41)^2

= 4 + 4√41 + 41

= 45 + 4√41

Therefore:

(x + 3)^2 + 45 + 4√41 = 81

So (x + 3)^2 = 81 - 45 - 4√41 = 36 - 4√41

Therefore, x + 3 = ±√(36 - 4√41)

Therefore, x = -3 ± √(36 - 4√41)

So the four points of intersection are:

For y = 10 + √41:

Point A: (-3 + √(36 + 4√41), 10 + √41)

Point B: (-3 - √(36 + 4√41), 10 + √41)

For y = 10 - √41:

Point C: (-3 + √(36 - 4√41), 10 - √41)

Point D: (-3 - √(36 - 4√41), 10 - √41)

So these are the four points. Now, I need to compute the sum of distances from each of these four points to (-3, 2). Let me denote the point (-3, 2) as P.

So, for each point, compute distance from P to that point, then add them all up.

Let me denote the four points as A, B, C, D.

Let me first compute the distance from P to A, then to B, then to C, then to D, and sum them.

But maybe there's a pattern or symmetry here. Let me think.

First, note that the center of the circle is at (-3, 12). The point P is at (-3, 2). So the line connecting the center of the circle to P is vertical, since both have x-coordinate -3. The distance between the center (-3, 12) and P (-3, 2) is |12 - 2| = 10 units.

Also, the radius of the circle is 9. So the circle is centered at (-3, 12) with radius 9, and P is inside the circle because the distance from center to P is 10, which is greater than the radius 9? Wait, 10 > 9, so P is outside the circle. Wait, but the problem says that the four points are intersections of the circle and hyperbola. So P is outside the circle. But maybe there's some relation?

Alternatively, maybe I can use coordinates. Let me think about the coordinates of the points. Let me consider points A and B. They have the same y-coordinate, 10 + √41, and x-coordinates symmetric around -3. Similarly for points C and D.

So for points A and B, their coordinates are (-3 ± a, y1), where a = √(36 + 4√41) and y1 = 10 + √41.

Similarly, points C and D are (-3 ± b, y2), where b = √(36 - 4√41) and y2 = 10 - √41.

So let me compute the distance from P (-3, 2) to each of these points.

First, distance from P to A:

sqrt[ ( (-3 + a) - (-3) )² + ( y1 - 2 )² ] = sqrt[ (a)² + (y1 - 2)² ]

Similarly, distance from P to B is sqrt[ (-a)² + (y1 - 2)² ] = same as above. So distance from P to A and P to B are equal. Similarly for C and D.

Therefore, the total sum is 2 * [ sqrt(a² + (y1 - 2)^2 ) + sqrt(b² + (y2 - 2)^2 ) ]

So let me compute these terms.

First, compute a² and b²:

From earlier:

For point A and B: a² = 36 + 4√41

For point C and D: b² = 36 - 4√41

Now, compute (y1 - 2) and (y2 - 2):

y1 = 10 + √41, so y1 - 2 = 8 + √41

y2 = 10 - √41, so y2 - 2 = 8 - √41

Therefore, the terms inside the square roots are:

For A and B: a² + (y1 - 2)^2 = (36 + 4√41) + (8 + √41)^2

Similarly for C and D: b² + (y2 - 2)^2 = (36 - 4√41) + (8 - √41)^2

Let me compute these expressions.

First, compute (8 + √41)^2:

= 8² + 2*8*√41 + (√41)^2

= 64 + 16√41 + 41

= 105 + 16√41

Similarly, (8 - √41)^2:

= 8² - 2*8*√41 + (√41)^2

= 64 - 16√41 + 41

= 105 - 16√41

Therefore, for A and B:

a² + (y1 - 2)^2 = 36 + 4√41 + 105 + 16√41 = (36 + 105) + (4√41 + 16√41) = 141 + 20√41

Similarly, for C and D:

b² + (y2 - 2)^2 = 36 - 4√41 + 105 - 16√41 = (36 + 105) + (-4√41 -16√41) = 141 - 20√41

Therefore, the distances from P to A and B are sqrt(141 + 20√41) and sqrt(141 - 20√41), each multiplied by 2 (since two points A and B). Wait, no. Wait, the total sum is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)].

Wait, no. Wait, for A and B, each distance is sqrt(141 + 20√41), so two of them, so 2*sqrt(141 + 20√41). Similarly for C and D, 2*sqrt(141 - 20√41). Wait, no. Wait, no. Wait, the total sum is:

For A and B: each distance is sqrt(141 + 20√41), so two points: 2*sqrt(141 + 20√41)

For C and D: each distance is sqrt(141 - 20√41), so two points: 2*sqrt(141 - 20√41)

Therefore, total sum is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41) = 2[ sqrt(141 + 20√41) + sqrt(141 - 20√41) ]

Hmm, this seems complicated. Maybe there's a way to simplify sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me denote S = sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me compute S².

S² = [sqrt(141 + 20√41) + sqrt(141 - 20√41)]²

= (sqrt(a) + sqrt(b))² = a + b + 2*sqrt(ab)

Where a = 141 + 20√41 and b = 141 - 20√41

Compute a + b = 141 + 20√41 + 141 - 20√41 = 282

Compute ab = (141)^2 - (20√41)^2

Compute 141²: Let me compute 140² + 2*140 + 1 = 19600 + 280 + 1 = 19881

Wait, 141² = (140 + 1)^2 = 140² + 2*140*1 + 1² = 19600 + 280 + 1 = 19881

Then (20√41)^2 = 400 * 41 = 16400

Therefore, ab = 19881 - 16400 = 3481

So sqrt(ab) = sqrt(3481). Let me check what is sqrt(3481). Let me think. 59² = 3481 because 60² = 3600, so 59² = 3481. Yes, because 60² - 2*60 + 1 = 3600 - 120 + 1 = 3481. So sqrt(ab) = 59.

Therefore, S² = 282 + 2*59 = 282 + 118 = 400

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

Therefore, the total sum is 2*S = 2*20 = 40? Wait, no. Wait, earlier I had total sum is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*S. But S was sqrt(...) + sqrt(...), which we found S = 20. Therefore, total sum is 2*20 = 40. Wait, but let me check again:

Wait, no. Wait, the total sum is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*S, where S = sqrt(...) + sqrt(...). But we found that S² = 400, so S = 20. Therefore, total sum is 2*20 = 40. Therefore, the answer is 40?

Wait, that seems too straightforward. Let me check again.

Let me recap:

Total sum of distances = 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*S, where S = sqrt(141 + 20√41) + sqrt(141 - 20√41). Then S² = 282 + 2*sqrt(ab) = 282 + 2*59 = 282 + 118 = 400. Therefore, S = 20. Therefore, total sum is 2*20 = 40. Therefore, the answer is 40. So the sum of distances is 40. Is that possible?

Alternatively, maybe there's a geometric interpretation. Let me think. The point (-3, 2) is located at (-3, 2). The circle is centered at (-3, 12) with radius 9. The distance between (-3, 2) and (-3, 12) is 10 units. The four points lie on the circle, so maybe there's some relation?

But the sum of distances from four points on a circle to a fixed point... Not sure. But according to the algebra, it's 40. Let me check with approximate values.

Let me compute approximate values for sqrt(141 + 20√41) and sqrt(141 - 20√41). Let me compute √41 ≈ 6.4031.

Compute 20√41 ≈ 20 * 6.4031 ≈ 128.062

Then 141 + 128.062 ≈ 269.062, sqrt(269.062) ≈ 16.4

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

Then adding them: 16.4 + 3.597 ≈ 19.997 ≈ 20. Then multiplying by 2 gives ≈ 40. So that checks out. Therefore, the exact value is 40. Therefore, the answer is 40.

But let me think again. Is there another way to see this? Maybe using properties of circles and hyperbolas?

Alternatively, since the four points lie on the circle centered at (-3, 12) with radius 9, and we are to find the sum of distances from (-3, 2) to each of these four points. Wait, but if we consider that the four points are on the circle, then maybe there's some relation?

But the sum of distances from a point to points on a circle... Not sure. However, in this case, the four points are not arbitrary points on the circle, but specific ones that also lie on the hyperbola. However, the algebra seems to confirm that the sum is 40. Also, the approximate calculation supports that. Therefore, I think the answer is 40.

But let me check once again my steps for errors.

First, subtracting equations to get y² - 20y + 59 = 0. Then solving for y gives 10 ± √41. Then substituting back into the circle equation to find x. Then computing distances. Then realizing that the sum simplifies to 40. Seems correct.

Alternatively, maybe there's a way to use coordinates. Let me think. Let me consider that the four points lie on the circle (x + 3)^2 + (y - 12)^2 = 81. Let me denote the point P as (-3, 2). Let me consider the distance from P to any point (x, y) on the circle. The distance squared is (x + 3)^2 + (y - 2)^2. But since (x + 3)^2 + (y - 12)^2 = 81, we can write (x + 3)^2 = 81 - (y - 12)^2. Therefore, distance squared is 81 - (y - 12)^2 + (y - 2)^2.

Compute that:

distance² = 81 - (y² - 24y + 144) + (y² - 4y + 4)

Simplify:

= 81 - y² + 24y - 144 + y² - 4y + 4

Simplify terms:

- y² + y² = 0

24y - 4y = 20y

81 - 144 + 4 = (81 + 4) - 144 = 85 - 144 = -59

Therefore, distance squared = 20y - 59

Therefore, distance = sqrt(20y - 59)

Wait, that's interesting! So for any point (x, y) on the circle, the distance squared to P(-3, 2) is 20y - 59. Therefore, the distance is sqrt(20y - 59). Therefore, the sum of distances from the four points to P is sum_{i=1}^4 sqrt(20y_i - 59), where y_i are the y-coordinates of the four points.

But earlier, we found that the y-coordinates are 10 ± √41 (each with multiplicity 2). Therefore, the four points have y-values: 10 + √41, 10 + √41, 10 - √41, 10 - √41.

Therefore, the sum is 2*sqrt(20*(10 + √41) - 59) + 2*sqrt(20*(10 - √41) - 59)

Compute inside the square roots:

For the first term: 20*(10 + √41) - 59 = 200 + 20√41 - 59 = 141 + 20√41

For the second term: 20*(10 - √41) - 59 = 200 - 20√41 - 59 = 141 - 20√41

Therefore, the sum is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41), which is exactly what I had before. Then, as before, this simplifies to 40. Therefore, this confirms the result.

Therefore, the answer is 40. Therefore, the sum of the distances is 40.

But let me check if the expression sqrt(20y - 59) is valid. For the distance to be real, 20y - 59 ≥ 0 => y ≥ 59/20 = 2.95. Let me check the y-values. The y-values are 10 ± √41 ≈ 10 ± 6.4, so approx 16.4 and 3.6. Both are greater than 2.95, so yes, the square roots are real. Therefore, this approach is valid.

Therefore, this method is much simpler. Instead of dealing with coordinates, I can use the relation derived from the circle equation. Therefore, the sum of distances is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 40. Therefore, the answer is 40.

Therefore, I think 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 told that these two curves intersect at four points, and we are to compute the **sum of the distances** from these four points to the point $(-3, 2)$.

---

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

- The first equation is a **circle** because the coefficients of $x^2$ and $y^2$ are equal and positive.
- The second equation is a **hyperbola** because the coefficients of $x^2$ and $y^2$ are of opposite signs.

Thus, the curves intersect at four points, as a circle and a hyperbola can intersect at up to four points.

---

### Step 2: Use Algebra to Find the Points of Intersection

We subtract the second equation from the first to eliminate $x^2$ and $6x$:

$$
(x^2 + y^2 + 6x - 24y + 72) - (x^2 - y^2 + 6x + 16y - 46) = 0
$$

Simplifying:

$$
2y^2 - 40y + 118 = 0 \Rightarrow y^2 - 20y + 59 = 0
$$

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 occurring twice (since the curves are symmetric in y).

---

### Step 3: Use the Circle Equation to Find x-Coordinates

We rewrite the first equation in standard form by completing the square:

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

Complete the square:

- $ x^2 + 6x = (x + 3)^2 - 9 $
- $ y^2 - 24y = (y - 12)^2 - 144 $

So the equation becomes:

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

This is a circle centered at $(-3, 12)$ with radius 9.

Now, for each $ y = 10 \pm \sqrt{41} $, we solve for $x$:

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

Let’s compute the values:

- For $ y = 10 + \sqrt{41} $, $ y - 12 = -2 + \sqrt{41} $, so:
  $$
  (x + 3)^2 = 81 - (-2 + \sqrt{41})^2 = 81 - (4 - 4\sqrt{41} + 41) = 36 + 4\sqrt{41}
  $$

- For $ y = 10 - \sqrt{41} $, $ y - 12 = -2 - \sqrt{41} $, so:
  $$
  (x + 3)^2 = 81 - (-2 - \sqrt{41})^2 = 81 - (4 + 4\sqrt{41} + 41) = 36 - 4\sqrt{41}
  $$

So the x-values are:

- $ x = -3 \pm \sqrt{36 + 4\sqrt{41}} $
- $ x = -3 \pm \sqrt{36 - 4\sqrt{41}} $

Thus, the four points of intersection are:

- $ (-3 + \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41}) $
- $ (-3 - \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41}) $
- $ (-3 + \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41}) $
- $ (-3 - \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41}) $

---

### Step 4: Compute the Sum of Distances to $(-3, 2)$

Let’s denote the point $ P = (-3, 2) $. The distance from $ P $ to a point $ (x, y) $ is:

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

From the circle equation, we know:

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

So the distance squared becomes:

$$
\text{Distance}^2 = 81 - (y - 12)^2 + (y - 2)^2
$$

Simplify:

$$
= 81 - (y^2 - 24y + 144) + (y^2 - 4y + 4)
= 81 - y^2 + 24y - 144 + y^2 - 4y + 4
= 20y - 59
$$

So the distance is:

$$
\text{Distance} = \sqrt{20y - 59}
$$

Now, the four points have y-values $ y = 10 \pm \sqrt{41} $, each occurring twice. So the total sum of distances is:

$$
2 \cdot \sqrt{20(10 + \sqrt{41}) - 59} + 2 \cdot \sqrt{20(10 - \sqrt{41}) - 59}
$$

Compute the expressions inside the square roots:

- $ 20(10 + \sqrt{41}) - 59 = 200 + 20\sqrt{41} - 59 = 141 + 20\sqrt{41} $
- $ 20(10 - \sqrt{41}) - 59 = 200 - 20\sqrt{41} - 59 = 141 - 20\sqrt{41} $

So the total sum is:

$$
2 \cdot \left( \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}} \right)
$$

Let’s denote:

$$
S = \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}}
$$

Then:

$$
S^2 = (141 + 20\sqrt{41}) + (141 - 20\sqrt{41}) + 2\sqrt{(141 + 20\sqrt{41})(141 - 20\sqrt{41})}
$$

$$
= 282 + 2\sqrt{141^2 - (20\sqrt{41})^2}
$$

$$
= 282 + 2\sqrt{19881 - 16400} = 282 + 2\sqrt{3481} = 282 + 2 \cdot 59 = 282 + 118 = 400
$$

So $ S = \sqrt{400} = 20 $

Therefore, the total sum of distances is:

$$
2 \cdot S = 2 \cdot 20 = \boxed{40}
$$

---

### 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 the intersection points might be complicated. Let me think if there's a smarter way than solving the equations directly. Maybe using some geometric properties?

Alternatively, maybe I can subtract the equations to eliminate some terms. Let me try that.

First, let me write both equations again:

Equation 1: $ x^2 + y^2 + 6x - 24y + 72 = 0 $

Equation 2: $ x^2 - y^2 + 6x + 16y - 46 = 0 $

If I subtract Equation 2 from Equation 1, I can eliminate the $ x^2 $ and $ 6x $ terms. Let me do that:

Equation 1 - Equation 2:

$ (x^2 + y^2 + 6x - 24y + 72) - (x^2 - y^2 + 6x + 16y - 46) = 0 - 0 $

Simplify term by term:

x² - x² = 0

y² - (-y²) = y² + y² = 2y²

6x - 6x = 0

-24y - 16y = -40y

72 - (-46) = 72 + 46 = 118

So altogether:

2y² - 40y + 118 = 0

Divide both sides by 2:

y² - 20y + 59 = 0

Wait, let me check that again. Wait, 72 + 46 is 118? 72 + 46 is 118? 70 + 40 is 110, 2 + 6 is 8, so 118. Yes. Then dividing by 2 gives y² - 20y + 59 = 0. Let me check the coefficients again.

Original subtraction:

Equation 1: + y², Equation 2: - y², so Equation 1 - Equation 2: + y² - (-y²) = 2y². Then -24y - 16y is -40y. Then +72 - (-46) is +72 + 46 = 118. So yes, 2y² - 40y + 118 = 0. Dividing by 2: y² - 20y + 59 = 0. Let me check discriminant: D = 400 - 4*1*59 = 400 - 236 = 164. Which is positive, so two real solutions. Therefore, there are two y-values where the curves intersect. But the problem says four points. So maybe each y corresponds to two x-values? But since one equation is a circle and the other is a hyperbola, maybe each y gives two x's? Let me check.

Alternatively, maybe I need to solve for y first. Let me solve the quadratic equation for y:

y² - 20y + 59 = 0

Using quadratic formula:

y = [20 ± sqrt(400 - 236)] / 2 = [20 ± sqrt(164)] / 2

Simplify sqrt(164): sqrt(4*41) = 2*sqrt(41). Therefore:

y = [20 ± 2√41]/2 = 10 ± √41

So two y-values: 10 + √41 and 10 - √41. So each of these y-values will correspond to some x-values. Let me find x-values for each y.

But to find x, I need to substitute back into one of the original equations. Let me pick Equation 1 or Equation 2? Maybe Equation 1 is a circle, so maybe easier to write in standard form?

Alternatively, maybe I can use Equation 2? Let me see. Let me try Equation 1 first.

Equation 1: $ x^2 + y^2 + 6x - 24y + 72 = 0 $

Let me rearrange this equation to standard form by completing the squares.

Group x terms and y terms:

x² + 6x + y² - 24y = -72

Complete the square for x:

x² + 6x = (x + 3)^2 - 9

For y:

y² - 24y = (y - 12)^2 - 144

So substituting back:

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

Combine constants:

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

Therefore:

(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 is $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Let me try to write this in standard form. Since it's a hyperbola, maybe I can rearrange terms.

Group x terms and y terms:

x² + 6x - y² + 16y = 46

Complete the square for x and y:

For x: x² + 6x = (x + 3)^2 - 9

For y: -y² + 16y = - (y² - 16y) = -[(y - 8)^2 - 64] = - (y - 8)^2 + 64

So substituting back:

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

Combine constants:

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

Therefore:

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

Which can be written as:

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

So this is a hyperbola centered at (-3, 8), with transverse axis along the y-axis, since the y-term is positive. The standard form is [(y - k)^2]/a² - [(x - h)^2]/b² = 1. Here, [(y - 8)^2]/9 - [(x + 3)^2]/9 = 1. Wait, because 9 is the right-hand side. So a² = 9, so a = 3, and b² = 9, so b = 3. Therefore, the hyperbola has asymptotes y - 8 = ±(b/a)(x + 3) => y - 8 = ±(x + 3). So asymptotes are y = x + 11 and y = -x + 5.

But maybe I don't need that. Anyway, the key point is that the hyperbola is centered at (-3, 8), and the circle is centered at (-3, 12). So both centers have the same x-coordinate, -3. So they are vertically aligned. The circle has radius 9, and the hyperbola is centered at (-3, 8). Let me check the distance between centers. The distance between (-3, 12) and (-3, 8) is |12 - 8| = 4 units. Since the circle has radius 9, which is larger than 4, so the hyperbola is inside the circle? Wait, but hyperbola extends infinitely. However, the intersection points are four points. So maybe the hyperbola intersects the circle at four points.

But going back, I need to find the points of intersection. Earlier, I found that the y-values are 10 ± √41. Let me compute approximate values to check if they make sense. √41 is approx 6.4, so 10 + 6.4 = 16.4, and 10 - 6.4 = 3.6. So y-values are around 16.4 and 3.6. The circle is centered at (-3, 12) with radius 9, so the topmost point of the circle is at y = 12 + 9 = 21, and bottommost at y = 12 - 9 = 3. So y = 16.4 is within the circle's range (since 16.4 < 21), and y = 3.6 is also within the circle's range (since 3.6 > 3). So that seems okay.

Now, to find x-values for each y. Let me use Equation 1 or Equation 2? Maybe Equation 2? Let me check. Let me use Equation 2: $ x^2 - y^2 + 6x + 16y - 46 = 0 $. Let me rearrange it as:

x² + 6x = y² - 16y + 46

But maybe it's better to use Equation 1. Let me try Equation 1 again. Since we have the circle equation in standard form, maybe we can use that. Let me recall that the circle equation is:

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

So if I know y, I can solve for x. Let me do that. Let me take the y-values we found: y = 10 ± √41.

Let me denote y = 10 + √41 first. Then plug into the circle equation:

(x + 3)^2 + ( (10 + √41) - 12 )^2 = 81

Simplify the y-term:

(10 + √41 - 12) = (-2 + √41)

Therefore:

(x + 3)^2 + (-2 + √41)^2 = 81

Compute (-2 + √41)^2:

= (-2)^2 + 2*(-2)*(√41) + (√41)^2

= 4 - 4√41 + 41

= 45 - 4√41

Therefore:

(x + 3)^2 + 45 - 4√41 = 81

So (x + 3)^2 = 81 - 45 + 4√41 = 36 + 4√41

Therefore, x + 3 = ±√(36 + 4√41)

Therefore, x = -3 ± √(36 + 4√41)

Similarly, for y = 10 - √41:

(x + 3)^2 + ( (10 - √41) - 12 )^2 = 81

Simplify the y-term:

(10 - √41 - 12) = (-2 - √41)

Therefore:

(x + 3)^2 + (-2 - √41)^2 = 81

Compute (-2 - √41)^2:

= (-2)^2 + 2*(-2)*(-√41) + (√41)^2

= 4 + 4√41 + 41

= 45 + 4√41

Therefore:

(x + 3)^2 + 45 + 4√41 = 81

So (x + 3)^2 = 81 - 45 - 4√41 = 36 - 4√41

Therefore, x + 3 = ±√(36 - 4√41)

Therefore, x = -3 ± √(36 - 4√41)

So the four points of intersection are:

For y = 10 + √41:

Point A: (-3 + √(36 + 4√41), 10 + √41)

Point B: (-3 - √(36 + 4√41), 10 + √41)

For y = 10 - √41:

Point C: (-3 + √(36 - 4√41), 10 - √41)

Point D: (-3 - √(36 - 4√41), 10 - √41)

So these are the four points. Now, I need to compute the sum of distances from each of these four points to (-3, 2). Let me denote the point (-3, 2) as P.

So, for each point, compute distance from P to that point, then add them all up.

Let me denote the four points as A, B, C, D.

Let me first compute the distance from P to A, then to B, then to C, then to D, and sum them.

But maybe there's a pattern or symmetry here. Let me think.

First, note that the center of the circle is at (-3, 12). The point P is at (-3, 2). So the line connecting the center of the circle to P is vertical, since both have x-coordinate -3. The distance between the center (-3, 12) and P (-3, 2) is |12 - 2| = 10 units.

Also, the radius of the circle is 9. So the circle is centered at (-3, 12) with radius 9, and P is inside the circle because the distance from center to P is 10, which is greater than the radius 9? Wait, 10 > 9, so P is outside the circle. Wait, but the problem says that the four points are intersections of the circle and hyperbola. So P is outside the circle. But maybe there's some relation?

Alternatively, maybe I can use coordinates. Let me think about the coordinates of the points. Let me consider points A and B. They have the same y-coordinate, 10 + √41, and x-coordinates symmetric around -3. Similarly for points C and D.

So for points A and B, their coordinates are (-3 ± a, y1), where a = √(36 + 4√41) and y1 = 10 + √41.

Similarly, points C and D are (-3 ± b, y2), where b = √(36 - 4√41) and y2 = 10 - √41.

So let me compute the distance from P (-3, 2) to each of these points.

First, distance from P to A:

sqrt[ ( (-3 + a) - (-3) )² + ( y1 - 2 )² ] = sqrt[ (a)² + (y1 - 2)² ]

Similarly, distance from P to B is sqrt[ (-a)² + (y1 - 2)² ] = same as above. So distance from P to A and P to B are equal. Similarly for C and D.

Therefore, the total sum is 2 * [ sqrt(a² + (y1 - 2)^2 ) + sqrt(b² + (y2 - 2)^2 ) ]

So let me compute these terms.

First, compute a² and b²:

From earlier:

For point A and B: a² = 36 + 4√41

For point C and D: b² = 36 - 4√41

Now, compute (y1 - 2) and (y2 - 2):

y1 = 10 + √41, so y1 - 2 = 8 + √41

y2 = 10 - √41, so y2 - 2 = 8 - √41

Therefore, the terms inside the square roots are:

For A and B: a² + (y1 - 2)^2 = (36 + 4√41) + (8 + √41)^2

Similarly for C and D: b² + (y2 - 2)^2 = (36 - 4√41) + (8 - √41)^2

Let me compute these expressions.

First, compute (8 + √41)^2:

= 8² + 2*8*√41 + (√41)^2

= 64 + 16√41 + 41

= 105 + 16√41

Similarly, (8 - √41)^2:

= 8² - 2*8*√41 + (√41)^2

= 64 - 16√41 + 41

= 105 - 16√41

Therefore, for A and B:

a² + (y1 - 2)^2 = 36 + 4√41 + 105 + 16√41 = (36 + 105) + (4√41 + 16√41) = 141 + 20√41

Similarly, for C and D:

b² + (y2 - 2)^2 = 36 - 4√41 + 105 - 16√41 = (36 + 105) + (-4√41 -16√41) = 141 - 20√41

Therefore, the distances from P to A and B are sqrt(141 + 20√41) and sqrt(141 - 20√41), each multiplied by 2 (since two points A and B). Wait, no. Wait, the total sum is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)].

Wait, no. Wait, for A and B, each distance is sqrt(141 + 20√41), so two of them, so 2*sqrt(141 + 20√41). Similarly for C and D, 2*sqrt(141 - 20√41). Wait, no. Wait, no. Wait, the total sum is:

For A and B: each distance is sqrt(141 + 20√41), so two points: 2*sqrt(141 + 20√41)

For C and D: each distance is sqrt(141 - 20√41), so two points: 2*sqrt(141 - 20√41)

Therefore, total sum is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41) = 2[ sqrt(141 + 20√41) + sqrt(141 - 20√41) ]

Hmm, this seems complicated. Maybe there's a way to simplify sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me denote S = sqrt(141 + 20√41) + sqrt(141 - 20√41). Let me compute S².

S² = [sqrt(141 + 20√41) + sqrt(141 - 20√41)]²

= (sqrt(a) + sqrt(b))² = a + b + 2*sqrt(ab)

Where a = 141 + 20√41 and b = 141 - 20√41

Compute a + b = 141 + 20√41 + 141 - 20√41 = 282

Compute ab = (141)^2 - (20√41)^2

Compute 141²: Let me compute 140² + 2*140 + 1 = 19600 + 280 + 1 = 19881

Wait, 141² = (140 + 1)^2 = 140² + 2*140*1 + 1² = 19600 + 280 + 1 = 19881

Then (20√41)^2 = 400 * 41 = 16400

Therefore, ab = 19881 - 16400 = 3481

So sqrt(ab) = sqrt(3481). Let me check what is sqrt(3481). Let me think. 59² = 3481 because 60² = 3600, so 59² = 3481. Yes, because 60² - 2*60 + 1 = 3600 - 120 + 1 = 3481. So sqrt(ab) = 59.

Therefore, S² = 282 + 2*59 = 282 + 118 = 400

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

Therefore, the total sum is 2*S = 2*20 = 40? Wait, no. Wait, earlier I had total sum is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*S. But S was sqrt(...) + sqrt(...), which we found S = 20. Therefore, total sum is 2*20 = 40. Wait, but let me check again:

Wait, no. Wait, the total sum is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*S, where S = sqrt(...) + sqrt(...). But we found that S² = 400, so S = 20. Therefore, total sum is 2*20 = 40. Therefore, the answer is 40?

Wait, that seems too straightforward. Let me check again.

Let me recap:

Total sum of distances = 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 2*S, where S = sqrt(141 + 20√41) + sqrt(141 - 20√41). Then S² = 282 + 2*sqrt(ab) = 282 + 2*59 = 282 + 118 = 400. Therefore, S = 20. Therefore, total sum is 2*20 = 40. Therefore, the answer is 40. So the sum of distances is 40. Is that possible?

Alternatively, maybe there's a geometric interpretation. Let me think. The point (-3, 2) is located at (-3, 2). The circle is centered at (-3, 12) with radius 9. The distance between (-3, 2) and (-3, 12) is 10 units. The four points lie on the circle, so maybe there's some relation?

But the sum of distances from four points on a circle to a fixed point... Not sure. But according to the algebra, it's 40. Let me check with approximate values.

Let me compute approximate values for sqrt(141 + 20√41) and sqrt(141 - 20√41). Let me compute √41 ≈ 6.4031.

Compute 20√41 ≈ 20 * 6.4031 ≈ 128.062

Then 141 + 128.062 ≈ 269.062, sqrt(269.062) ≈ 16.4

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

Then adding them: 16.4 + 3.597 ≈ 19.997 ≈ 20. Then multiplying by 2 gives ≈ 40. So that checks out. Therefore, the exact value is 40. Therefore, the answer is 40.

But let me think again. Is there another way to see this? Maybe using properties of circles and hyperbolas?

Alternatively, since the four points lie on the circle centered at (-3, 12) with radius 9, and we are to find the sum of distances from (-3, 2) to each of these four points. Wait, but if we consider that the four points are on the circle, then maybe there's some relation?

But the sum of distances from a point to points on a circle... Not sure. However, in this case, the four points are not arbitrary points on the circle, but specific ones that also lie on the hyperbola. However, the algebra seems to confirm that the sum is 40. Also, the approximate calculation supports that. Therefore, I think the answer is 40.

But let me check once again my steps for errors.

First, subtracting equations to get y² - 20y + 59 = 0. Then solving for y gives 10 ± √41. Then substituting back into the circle equation to find x. Then computing distances. Then realizing that the sum simplifies to 40. Seems correct.

Alternatively, maybe there's a way to use coordinates. Let me think. Let me consider that the four points lie on the circle (x + 3)^2 + (y - 12)^2 = 81. Let me denote the point P as (-3, 2). Let me consider the distance from P to any point (x, y) on the circle. The distance squared is (x + 3)^2 + (y - 2)^2. But since (x + 3)^2 + (y - 12)^2 = 81, we can write (x + 3)^2 = 81 - (y - 12)^2. Therefore, distance squared is 81 - (y - 12)^2 + (y - 2)^2.

Compute that:

distance² = 81 - (y² - 24y + 144) + (y² - 4y + 4)

Simplify:

= 81 - y² + 24y - 144 + y² - 4y + 4

Simplify terms:

- y² + y² = 0

24y - 4y = 20y

81 - 144 + 4 = (81 + 4) - 144 = 85 - 144 = -59

Therefore, distance squared = 20y - 59

Therefore, distance = sqrt(20y - 59)

Wait, that's interesting! So for any point (x, y) on the circle, the distance squared to P(-3, 2) is 20y - 59. Therefore, the distance is sqrt(20y - 59). Therefore, the sum of distances from the four points to P is sum_{i=1}^4 sqrt(20y_i - 59), where y_i are the y-coordinates of the four points.

But earlier, we found that the y-coordinates are 10 ± √41 (each with multiplicity 2). Therefore, the four points have y-values: 10 + √41, 10 + √41, 10 - √41, 10 - √41.

Therefore, the sum is 2*sqrt(20*(10 + √41) - 59) + 2*sqrt(20*(10 - √41) - 59)

Compute inside the square roots:

For the first term: 20*(10 + √41) - 59 = 200 + 20√41 - 59 = 141 + 20√41

For the second term: 20*(10 - √41) - 59 = 200 - 20√41 - 59 = 141 - 20√41

Therefore, the sum is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41), which is exactly what I had before. Then, as before, this simplifies to 40. Therefore, this confirms the result.

Therefore, the answer is 40. Therefore, the sum of the distances is 40.

But let me check if the expression sqrt(20y - 59) is valid. For the distance to be real, 20y - 59 ≥ 0 => y ≥ 59/20 = 2.95. Let me check the y-values. The y-values are 10 ± √41 ≈ 10 ± 6.4, so approx 16.4 and 3.6. Both are greater than 2.95, so yes, the square roots are real. Therefore, this approach is valid.

Therefore, this method is much simpler. Instead of dealing with coordinates, I can use the relation derived from the circle equation. Therefore, the sum of distances is 2*[sqrt(141 + 20√41) + sqrt(141 - 20√41)] = 40. Therefore, the answer is 40.

Therefore, I think 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 told that these two curves intersect at four points, and we are to compute the **sum of the distances** from these four points to the point $(-3, 2)$.

---

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

- The first equation is a **circle** because the coefficients of $x^2$ and $y^2$ are equal and positive.
- The second equation is a **hyperbola** because the coefficients of $x^2$ and $y^2$ are of opposite signs.

Thus, the curves intersect at four points, as a circle and a hyperbola can intersect at up to four points.

---

### Step 2: Use Algebra to Find the Points of Intersection

We subtract the second equation from the first to eliminate $x^2$ and $6x$:

$$
(x^2 + y^2 + 6x - 24y + 72) - (x^2 - y^2 + 6x + 16y - 46) = 0
$$

Simplifying:

$$
2y^2 - 40y + 118 = 0 \Rightarrow y^2 - 20y + 59 = 0
$$

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 occurring twice (since the curves are symmetric in y).

---

### Step 3: Use the Circle Equation to Find x-Coordinates

We rewrite the first equation in standard form by completing the square:

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

Complete the square:

- $ x^2 + 6x = (x + 3)^2 - 9 $
- $ y^2 - 24y = (y - 12)^2 - 144 $

So the equation becomes:

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

This is a circle centered at $(-3, 12)$ with radius 9.

Now, for each $ y = 10 \pm \sqrt{41} $, we solve for $x$:

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

Let’s compute the values:

- For $ y = 10 + \sqrt{41} $, $ y - 12 = -2 + \sqrt{41} $, so:
  $$
  (x + 3)^2 = 81 - (-2 + \sqrt{41})^2 = 81 - (4 - 4\sqrt{41} + 41) = 36 + 4\sqrt{41}
  $$

- For $ y = 10 - \sqrt{41} $, $ y - 12 = -2 - \sqrt{41} $, so:
  $$
  (x + 3)^2 = 81 - (-2 - \sqrt{41})^2 = 81 - (4 + 4\sqrt{41} + 41) = 36 - 4\sqrt{41}
  $$

So the x-values are:

- $ x = -3 \pm \sqrt{36 + 4\sqrt{41}} $
- $ x = -3 \pm \sqrt{36 - 4\sqrt{41}} $

Thus, the four points of intersection are:

- $ (-3 + \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41}) $
- $ (-3 - \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41}) $
- $ (-3 + \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41}) $
- $ (-3 - \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41}) $

---

### Step 4: Compute the Sum of Distances to $(-3, 2)$

Let’s denote the point $ P = (-3, 2) $. The distance from $ P $ to a point $ (x, y) $ is:

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

From the circle equation, we know:

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

So the distance squared becomes:

$$
\text{Distance}^2 = 81 - (y - 12)^2 + (y - 2)^2
$$

Simplify:

$$
= 81 - (y^2 - 24y + 144) + (y^2 - 4y + 4)
= 81 - y^2 + 24y - 144 + y^2 - 4y + 4
= 20y - 59
$$

So the distance is:

$$
\text{Distance} = \sqrt{20y - 59}
$$

Now, the four points have y-values $ y = 10 \pm \sqrt{41} $, each occurring twice. So the total sum of distances is:

$$
2 \cdot \sqrt{20(10 + \sqrt{41}) - 59} + 2 \cdot \sqrt{20(10 - \sqrt{41}) - 59}
$$

Compute the expressions inside the square roots:

- $ 20(10 + \sqrt{41}) - 59 = 200 + 20\sqrt{41} - 59 = 141 + 20\sqrt{41} $
- $ 20(10 - \sqrt{41}) - 59 = 200 - 20\sqrt{41} - 59 = 141 - 20\sqrt{41} $

So the total sum is:

$$
2 \cdot \left( \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}} \right)
$$

Let’s denote:

$$
S = \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}}
$$

Then:

$$
S^2 = (141 + 20\sqrt{41}) + (141 - 20\sqrt{41}) + 2\sqrt{(141 + 20\sqrt{41})(141 - 20\sqrt{41})}
$$

$$
= 282 + 2\sqrt{141^2 - (20\sqrt{41})^2}
$$

$$
= 282 + 2\sqrt{19881 - 16400} = 282 + 2\sqrt{3481} = 282 + 2 \cdot 59 = 282 + 118 = 400
$$

So $ S = \sqrt{400} = 20 $

Therefore, the total sum of distances is:

$$
2 \cdot S = 2 \cdot 20 = \boxed{40}
$$

---

### 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 $2y_1 + 2y_2 = \boxed{40}.$