Explanation:
Okay, so I need to find the sum of the distances from the four intersection points of these two circles (or maybe one is a hyperbola?) to the point (-3, 2). Let me start by understanding the equations given.

First, the equations are:

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

Hmm, the first one is a quadratic equation in x and y. Let me check if they are circles or something else. For a circle, the general equation is $ x^2 + y^2 + Dx + Ey + F = 0 $. So the first equation is a circle. The second equation has $ x^2 - y^2 $, which makes me think it might be a hyperbola. Let me confirm that.

For the first equation: $ x^2 + y^2 + 6x - 24y + 72 = 0 $. To convert this to standard form, I need to complete the squares for x and y.

Starting with the first equation:

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 terms: 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 this is a circle with center at (-3, 12) and radius √81 = 9. Okay, so that's a circle.

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

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

Hmm, maybe complete the square for x and y. Let's see.

For x terms: $ x^2 + 6x $. Complete square: as before, add and subtract 9.

For y terms: -y^2 + 16y. Let me factor out the negative sign first: - (y^2 - 16y). Then complete the square inside the parentheses. Take coefficient of y, -16, divide by 2: -8, square it: 64. So:

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

So putting back into the equation:

Original equation after grouping:

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

Which becomes:

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

So:

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

Simplify constants:

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

Subtract 55 from both sides:

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

Which can be written as:

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

This is a hyperbola. So the first equation is a circle, the second is a hyperbola. Their intersection points are four points. Wait, but hyperbola and circle can intersect at up to four points. So yes, four points.

But the problem says they intersect at four points. So I need to find those four points, compute their distances to (-3, 2), and sum them up.

But computing four points and then calculating distance to (-3, 2) might be tedious. Maybe there's a smarter way?

Alternatively, maybe there is some geometric interpretation. Let me think.

First, note that (-3, 2) is the point... Let me check: The center of the first circle is at (-3, 12). So (-3, 2) is directly below the center of the first circle. The distance between (-3, 2) and (-3, 12) is 10 units. Since the radius of the first circle is 9, so (-3, 2) is inside the first circle because 10 > 9? Wait, distance from (-3, 2) to center (-3, 12) is 10 units. But radius is 9, so (-3, 2) is outside the first circle? Wait, distance from center to the point is 10, which is greater than radius 9, so yes, it's outside. So the point (-3, 2) is outside the first circle. However, since the second equation is a hyperbola, which is different.

But maybe there is a way to relate the distances?

Alternatively, maybe the problem can be approached by solving the equations and then computing the distances. Let me try to find the intersection points.

We have two equations:

1. $ (x + 3)^2 + (y - 12)^2 = 81 $ (Circle C1)
2. $ (y - 8)^2 - (x + 3)^2 = 9 $ (Hyperbola H)

So perhaps we can solve these two equations simultaneously. Let me denote:

From equation 1: Let me expand equation 1 again to see if I can use substitution with equation 2. Wait, maybe not. Let me note that equation 2 can be rearranged as (y - 8)^2 = (x + 3)^2 + 9.

So let me substitute this into equation 1. Equation 1 is:

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

But (y - 12) can be expressed in terms of (y - 8). Let me compute (y - 12) = (y - 8) - 4. So:

$ (x + 3)^2 + [(y - 8) - 4]^2 = 81 $

Which is:

$ (x + 3)^2 + (y - 8)^2 - 8(y - 8) + 16 = 81 $

But from equation 2, we know that (y - 8)^2 = (x + 3)^2 + 9. Let me substitute that here.

So replacing (y - 8)^2:

$ (x + 3)^2 + [(x + 3)^2 + 9] - 8(y - 8) + 16 = 81 $

Combine terms:

First, expand:

$ (x + 3)^2 + (x + 3)^2 + 9 - 8(y - 8) + 16 = 81 $

Combine like terms:

2*(x + 3)^2 + 25 - 8(y - 8) = 81

Subtract 25 from both sides:

2*(x + 3)^2 - 8(y - 8) = 56

Divide both sides by 2:

(x + 3)^2 - 4(y - 8) = 28

Hmm, not sure if helpful yet. Let me check if I can express y in terms of x or vice versa. Alternatively, maybe use equation 2 to express (x + 3)^2.

From equation 2: (y - 8)^2 - (x + 3)^2 = 9 => (x + 3)^2 = (y - 8)^2 - 9. Let me plug that into the previous equation.

So the equation above was:

(x + 3)^2 - 4(y - 8) = 28. Substitute (x + 3)^2:

(y - 8)^2 - 9 - 4(y - 8) = 28

Let me denote z = y - 8. Then equation becomes:

z^2 - 9 - 4z = 28 => z^2 - 4z - 37 = 0

Solving quadratic equation: z = [4 ± sqrt(16 + 148)] / 2 = [4 ± sqrt(164)] / 2 = [4 ± 2*sqrt(41)] / 2 = 2 ± sqrt(41)

Therefore, z = 2 + sqrt(41) or z = 2 - sqrt(41). Since z = y - 8, then y = 8 + z = 8 + 2 ± sqrt(41) = 10 ± sqrt(41)

So y = 10 + sqrt(41) and y = 10 - sqrt(41). Now, let's find corresponding x values.

From equation 2: (y - 8)^2 - (x + 3)^2 = 9 => (x + 3)^2 = (y - 8)^2 - 9

Since we have y, let's compute (y - 8)^2 for each case.

For y = 10 + sqrt(41):

(y - 8) = 2 + sqrt(41), so (y - 8)^2 = (2 + sqrt(41))^2 = 4 + 4*sqrt(41) + 41 = 45 + 4*sqrt(41)

Then, (x + 3)^2 = 45 + 4*sqrt(41) - 9 = 36 + 4*sqrt(41)

Similarly, for y = 10 - sqrt(41):

(y - 8) = 2 - sqrt(41), so (y - 8)^2 = (2 - sqrt(41))^2 = 4 - 4*sqrt(41) + 41 = 45 - 4*sqrt(41)

Then, (x + 3)^2 = 45 - 4*sqrt(41) - 9 = 36 - 4*sqrt(41)

So for each y value, we have two x values.

So for y = 10 + sqrt(41):

(x + 3)^2 = 36 + 4*sqrt(41). Therefore, x + 3 = ± sqrt(36 + 4*sqrt(41)), so x = -3 ± sqrt(36 + 4*sqrt(41))

Similarly, for y = 10 - sqrt(41):

(x + 3)^2 = 36 - 4*sqrt(41). Therefore, sqrt(36 - 4*sqrt(41))... Wait, is 36 - 4*sqrt(41) positive?

Compute sqrt(41) ≈ 6.403, so 4*sqrt(41) ≈ 25.612, so 36 - 25.612 ≈ 10.388, which is positive. Therefore, sqrt(36 - 4*sqrt(41)) is real, so x = -3 ± sqrt(36 - 4*sqrt(41))

Therefore, the four points of intersection are:

1. (-3 + sqrt(36 + 4*sqrt(41)), 10 + sqrt(41))
2. (-3 - sqrt(36 + 4*sqrt(41)), 10 + sqrt(41))
3. (-3 + sqrt(36 - 4*sqrt(41)), 10 - sqrt(41))
4. (-3 - sqrt(36 - 4*sqrt(41)), 10 - sqrt(41))

That seems complicated. Now, I need to compute the sum of distances from each of these four points to (-3, 2).

Let me denote the four points as P1, P2, P3, P4.

Let me note that each point has x-coordinate of the form -3 ± something, and y-coordinate of 10 ± sqrt(41). Let me think about the distance formula.

For a general point (x, y), distance to (-3, 2) is sqrt[(x + 3)^2 + (y - 2)^2]

For each point:

For P1: x = -3 + sqrt(36 + 4*sqrt(41)), y = 10 + sqrt(41)

So distance squared is [sqrt(36 + 4*sqrt(41))]^2 + [ (10 + sqrt(41) - 2) ]^2 = (36 + 4*sqrt(41)) + [8 + sqrt(41)]^2

Similarly for P2: x = -3 - sqrt(...), so [sqrt(...)]^2 is same as above, but since it's negative, when squared it's same. Therefore, distance squared for P2 is same as P1. Similarly for P3 and P4.

Therefore, the sum of distances squared would be 2*(distance squared of P1 + distance squared of P3). However, the problem asks for the sum of distances, not squared distances. Hmm. But maybe there's something with symmetry?

Alternatively, perhaps the distances have some relation. Let me think.

Alternatively, maybe we can compute the sum of distances by using properties of circles and hyperbola.

Wait, but the point (-3, 2) is interesting. Let me check. The center of the first circle is (-3, 12). The point (-3, 2) is directly below that center. Also, the equation of the first circle is (x + 3)^2 + (y - 12)^2 = 81. So, (-3, 2) is a point outside of this circle because the distance between them is 10, which is larger than the radius 9.

But let me think: If we consider the four points of intersection between the circle and hyperbola. Maybe they lie on some relation with respect to (-3, 2). Let me consider that.

Alternatively, maybe I can use parametrization or coordinates. Let me try to compute the distances.

First, for P1 and P2, their x-coordinates are symmetric around -3, and their y-coordinate is same. Similarly for P3 and P4.

Let me compute the distance for P1:

Distance from (-3, 2) to P1:

sqrt[ (x - (-3))^2 + (y - 2)^2 ] = sqrt[ (x + 3)^2 + (y - 2)^2 ]

But x + 3 is sqrt(36 + 4√41) or -sqrt(...), but squared it's same. So for P1 and P2, distance squared is:

(x + 3)^2 + (y - 2)^2 = (36 + 4√41) + (10 + √41 - 2)^2 = 36 + 4√41 + (8 + √41)^2

Similarly, compute (8 + √41)^2:

8^2 + 2*8*√41 + (√41)^2 = 64 + 16√41 + 41 = 105 + 16√41

Therefore, total distance squared for P1 and P2 combined:

2*(36 + 4√41 + 105 + 16√41) = 2*(141 + 20√41) ?

Wait, wait. Wait, no: For each of P1 and P2, their distance squared is 36 + 4√41 + 105 + 16√41? Wait no, that would be if P1 and P2 have same distance squared. Wait, let me redo:

For P1: distance squared is (x + 3)^2 + (y - 2)^2 = [sqrt(36 + 4√41)]^2 + [ (10 + √41 - 2) ]^2 = 36 + 4√41 + (8 + √41)^2. As above, which is 36 + 4√41 + 105 + 16√41 = 36 + 105 + 4√41 + 16√41 = 141 + 20√41.

Similarly for P2, same value. Therefore, sum of distances squared for P1 and P2 is 2*(141 + 20√41). Similarly for P3 and P4.

Let me compute for P3 and P4:

For P3, y = 10 - sqrt(41), so (y - 2) = 8 - sqrt(41). Then, (y - 2)^2 = (8 - sqrt(41))^2 = 64 - 16√41 + 41 = 105 - 16√41.

And (x + 3)^2 is 36 - 4√41. Therefore, distance squared for P3 is:

36 - 4√41 + 105 - 16√41 = 141 - 20√41. Similarly for P4. So total distance squared for P3 and P4 is 2*(141 - 20√41).

Therefore, total sum of squared distances is 2*(141 + 20√41) + 2*(141 - 20√41) = 2*141 + 2*20√41 + 2*141 - 2*20√41 = 4*141 = 564.

But the problem asks for the sum of distances, not the sum of squared distances. However, maybe there is a way to relate the sum of distances to the sum of squared distances? Not sure. Alternatively, maybe we can consider complex numbers or something else?

Alternatively, perhaps there is a geometric approach. Let me think again.

Wait, if I think about the sum of distances from four points to a fixed point. But unless there's some symmetry, it's difficult. But let me think about the four points.

Note that the first circle has center (-3, 12) and radius 9. The point (-3, 2) is along the vertical line x = -3, 10 units below the center.

Also, the hyperbola equation is (y - 8)^2 - (x + 3)^2 = 9. Which is a hyperbola centered at (-3, 8). So the center of the hyperbola is at (-3, 8). The circle is centered at (-3, 12), which is 4 units above the hyperbola's center.

But how does this help? Maybe not directly.

Alternatively, maybe think of the four points as lying on both the circle and the hyperbola. Let me check if these points can be related to (-3, 2).

Alternatively, maybe use complex numbers or parametric equations. Alternatively, think about inversion? Not sure.

Alternatively, since the problem is asking for the sum of distances, and we have expressions for the squared distances. Maybe we can compute the sum of square roots?

But that seems complicated. However, note that if we have sum of distances, and we know sum of squared distances, but unless we have more info, it's not helpful. But maybe there is a way.

Alternatively, perhaps the four points lie on a circle with diameter between (-3, 2) and something? Not sure.

Alternatively, think about Vieta's formula. Wait, maybe if we consider the four points on both curves, and relate to coordinates. But maybe not straightforward.

Wait, let me go back. Earlier, I calculated the sum of squared distances to (-3, 2) as 564. But maybe there's a way to compute the sum of distances?

Alternatively, maybe note that the four points lie on the circle C1. Let me check.

Yes, all four points lie on the circle C1. So, for each point (x, y) on C1, we have (x + 3)^2 + (y - 12)^2 = 81.

Let me denote P = (-3, 2). Let me compute the distance from P to any point on C1.

But the sum of distances from four points on the circle to P... Not sure. But maybe there's some relation. Wait, but the four points are not arbitrary on the circle. They also lie on the hyperbola.

Alternatively, maybe parametrize the circle and find points also on the hyperbola? But that seems similar to what I did before.

Wait, here's an idea. Since the points lie on both the circle and hyperbola, maybe I can find a relation between (x + 3)^2 and (y - 8)^2 from the hyperbola equation.

From hyperbola equation: (y - 8)^2 - (x + 3)^2 = 9 => (x + 3)^2 = (y - 8)^2 - 9.

From circle equation: (x + 3)^2 + (y - 12)^2 = 81. Substitute (x + 3)^2:

(y - 8)^2 - 9 + (y - 12)^2 = 81 => (y - 8)^2 + (y - 12)^2 = 90.

Expanding both squares:

(y² - 16y + 64) + (y² - 24y + 144) = 90

Combine terms:

2y² - 40y + 208 = 90 => 2y² - 40y + 118 = 0 => Divide by 2: y² - 20y + 59 = 0

Wait, earlier when solving for y, we had y = 10 ± sqrt(41). Let me check if that satisfies y² - 20y + 59 = 0.

Let me compute for y = 10 + sqrt(41):

y² = (10 + sqrt(41))² = 100 + 20*sqrt(41) + 41 = 141 + 20*sqrt(41)

Then y² - 20y + 59 = (141 + 20√41) - 20*(10 + √41) + 59 = 141 + 20√41 - 200 - 20√41 + 59 = (141 - 200 + 59) + (20√41 - 20√41) = 0. Correct. Similarly for the other root.

So, this gives us y² - 20y + 59 = 0.

So for each of the four points, their y-coordinates satisfy this quadratic. Also, the x-coordinates can be found using the earlier equations. But how does this help?

Alternatively, think about the distances from (-3, 2) to each point. Let me denote D as the distance from (-3, 2) to a point (x, y). Then, D² = (x + 3)² + (y - 2)².

But from the circle equation, (x + 3)² = 81 - (y - 12)². Therefore, D² = 81 - (y - 12)² + (y - 2)².

Compute that:

D² = 81 - [ (y² - 24y + 144) - (y² - 4y + 4) ]? Wait, wait:

Wait, (y - 12)^2 is y² - 24y + 144, and (y - 2)^2 is y² - 4y + 4. Therefore:

D² = 81 - [ (y² - 24y + 144) ] + (y² - 4y + 4) = 81 - y² + 24y - 144 + y² - 4y + 4.

Simplify terms:

- y² + y² cancels.

Then 81 - 144 + 4 + 24y - 4y = (-59) + 20y. So D² = 20y - 59.

Wow! Therefore, for each of the four points, D² = 20y - 59. Therefore, D = sqrt(20y - 59).

But since we have four points, each with y-coordinate being 10 ± sqrt(41), let me check that.

For the two values of y:

First case: y = 10 + sqrt(41). Then D² = 20*(10 + sqrt(41)) - 59 = 200 + 20*sqrt(41) - 59 = 141 + 20*sqrt(41).

Which matches with what I had earlier. Similarly, for y = 10 - sqrt(41): D² = 20*(10 - sqrt(41)) - 59 = 200 - 20*sqrt(41) - 59 = 141 - 20*sqrt(41).

So, for each of the four points, their distances squared are 141 ± 20√41 (each occurring twice).

But the problem asks for the sum of distances, not squared distances. However, perhaps I can find the sum of sqrt(20y - 59) for each of the four points.

But since two of them have y = 10 + sqrt(41) and two have y = 10 - sqrt(41), then sum of distances is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41).

So, sum = 2[ sqrt(141 + 20√41) + sqrt(141 - 20√41) ]

Hmm, need to compute this expression. Let me denote A = sqrt(141 + 20√41) + sqrt(141 - 20√41). Then sum is 2A. Let me compute A.

Let me compute A²:

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

= (sqrt(a) + sqrt(b))² where a = 141 + 20√41, b = 141 - 20√41

= a + b + 2*sqrt(ab)

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

Compute ab = (141 + 20√41)(141 - 20√41) = 141² - (20√41)²

Calculate:

141²: Well, 140² = 19600, 140² + 2*140 + 1 = 19600 + 280 + 1 = 19881.

(20√41)² = 400 * 41 = 16400

So ab = 19881 - 16400 = 3481.

So sqrt(ab) = sqrt(3481). Let me check sqrt(3481).

Well, 59² = 3481 because 60²=3600, so 59² = 3481. Yes. Therefore, sqrt(ab) = 59.

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

Therefore, A = sqrt(400) = 20.

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

Wow! So the answer is 40.

But let me just check my steps again to make sure I didn't make a mistake.

First, I found that D² = 20y - 59 for each point. Then, since y for two points is 10 + sqrt(41) and two points 10 - sqrt(41), so each of their D squared is 20*(10 ± sqrt(41)) - 59. Then, the sum is 2*sqrt(20*(10 + sqrt(41)) -59) + 2*sqrt(...).

Then, I computed that expression and found that A² = 400, so A = 20, so total sum is 40. Seems correct.

Alternatively, maybe there's a smarter way to see that the sum is 40.

Alternatively, note that when I found D² = 20y - 59, then for each point on the circle, since two points have y = 10 + sqrt(41) and two have y = 10 - sqrt(41), so total sum of D is 2*(sqrt(20*(10 + sqrt(41)) -59) + sqrt(20*(10 - sqrt(41)) -59)) )

But then, simplifying inside the square roots:

20*(10 + sqrt(41)) -59 = 200 + 20√41 -59 = 141 + 20√41

Similarly for the other term. Then, we found that sqrt(141 + 20√41) + sqrt(141 - 20√41) = 20.

Therefore, total sum is 2*20 = 40.

Therefore, the answer is 40.

Alternatively, maybe there is a way to see that the sum of sqrt(141 + 20√41) + sqrt(141 - 20√41) equals 20? Let me check with numbers.

Let me compute sqrt(141 + 20√41) + sqrt(141 - 20√41). Suppose this equals 20. Then, let me compute sqrt(141 + 20√41). Let me approximate values.

First, sqrt(41) ≈ 6.403. Then 20*sqrt(41) ≈ 128.06. Therefore, 141 + 128.06 ≈ 269.06. sqrt(269.06) ≈ 16.4.

Similarly, 141 - 128.06 ≈ 12.94. sqrt(12.94) ≈ 3.597. Then 16.4 + 3.597 ≈ 19.997 ≈ 20. So yes, it checks out numerically.

Therefore, the answer is 40.

Therefore, I think that the answer is 40.

**Final Answer**
\boxed{40}

To solve the problem, we are given two conic equations:

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

The goal is to compute the **sum of the distances** from the four intersection points of these conics to the point $(-3, 2)$.

---

### Step 1: Recognize the Shapes of the Curves

- The first equation is a **circle**, and after completing the square:
  $$
  (x + 3)^2 + (y - 12)^2 = 81
  $$
  This is a circle with **center at (-3, 12)** and **radius 9**.

- The second equation is a **hyperbola**, and after rearranging:
  $$
  (y - 8)^2 - (x + 3)^2 = 9
  $$
  This is a vertical hyperbola centered at $(-3, 8)$, opening up and down.

---

### Step 2: Find Coordinates of the Intersection Points

By solving both equations simultaneously, we find that the four points of intersection lie at:

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

This gives us four distinct points of intersection.

---

### Step 3: Compute the Sum of Distances

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

Let’s define the distance from a point $ (x, y) $ to $(-3, 2)$ as:
$$
D = \sqrt{(x + 3)^2 + (y - 2)^2}
$$

We simplify using the fact that all four points lie on the circle centered at $(-3, 12)$, so we can substitute the circle's equation:
$$
(x + 3)^2 + (y - 12)^2 = 81 \Rightarrow (x + 3)^2 = 81 - (y - 12)^2
$$

Then the square of the distance becomes:
$$
D^2 = (x + 3)^2 + (y - 2)^2 = 81 - (y - 12)^2 + (y - 2)^2
$$

Simplifying:
$$
D^2 = 81 - [(y - 12)^2 - (y - 2)^2]
$$

Expanding both squared terms:
$$
(y - 12)^2 = y^2 - 24y + 144,\quad (y - 2)^2 = y^2 - 4y + 4
$$
$$
D^2 = 81 - [y^2 - 24y + 144 - (y^2 - 4y + 4)] = 81 - [-20y + 140] = 81 + 20y - 140 = 20y - 59
$$

Therefore, the distance from each point to (-3, 2) is:
$$
D = \sqrt{20y - 59}
$$

---

### Step 4: Use Symmetry and Roots

From earlier steps, we found that the **y-coordinates** of the four points satisfy:
$$
y^2 - 20y + 59 = 0
$$
Thus, the two distinct y-values are:
$$
y = 10 \pm \sqrt{41}
$$

So the four distance contributions are:
- Two points with $ y = 10 + \sqrt{41} \Rightarrow D = \sqrt{20(10 + \sqrt{41}) - 59} = \sqrt{141 + 20\sqrt{41}} $
- Two points with $ y = 10 - \sqrt{41} \Rightarrow D = \sqrt{20(10 - \sqrt{41}) - 59} = \sqrt{141 - 20\sqrt{41}} $

Therefore, the **sum of distances** is:
$$
\text{Sum} = 2\left( \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}} \right)
$$

Let us denote:
$$
A = \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}}
$$

We square $A$:
$$
A^2 = \left( \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}} \right)^2
$$
$$
= (141 + 20\sqrt{41}) + (141 - 20\sqrt{41}) + 2\sqrt{(141 + 20\sqrt{41})(141 - 20\sqrt{41})}
$$

Simplify:
- $141 + 20\sqrt{41} + 141 - 20\sqrt{41} = 282$
- $(141 + 20\sqrt{41})(141 - 20\sqrt{41}) = 141^2 - (20\sqrt{41})^2 = 19881 - 16400 = 3481$

So:
$$
A^2 = 282 + 2\sqrt{3481} = 282 + 2 \cdot 59 = 282 + 118 = 400
\Rightarrow A = \sqrt{400} = 20
$$

Hence, the total distance is:
$$
2A = 2 \cdot 20 = \boxed{40}
$$

---

### 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 (or maybe one is a hyperbola?) to the point (-3, 2). Let me start by understanding the equations given.

First, the equations are:

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

Hmm, the first one is a quadratic equation in x and y. Let me check if they are circles or something else. For a circle, the general equation is $ x^2 + y^2 + Dx + Ey + F = 0 $. So the first equation is a circle. The second equation has $ x^2 - y^2 $, which makes me think it might be a hyperbola. Let me confirm that.

For the first equation: $ x^2 + y^2 + 6x - 24y + 72 = 0 $. To convert this to standard form, I need to complete the squares for x and y.

Starting with the first equation:

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 terms: 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 this is a circle with center at (-3, 12) and radius √81 = 9. Okay, so that's a circle.

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

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

Hmm, maybe complete the square for x and y. Let's see.

For x terms: $ x^2 + 6x $. Complete square: as before, add and subtract 9.

For y terms: -y^2 + 16y. Let me factor out the negative sign first: - (y^2 - 16y). Then complete the square inside the parentheses. Take coefficient of y, -16, divide by 2: -8, square it: 64. So:

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

So putting back into the equation:

Original equation after grouping:

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

Which becomes:

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

So:

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

Simplify constants:

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

Subtract 55 from both sides:

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

Which can be written as:

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

This is a hyperbola. So the first equation is a circle, the second is a hyperbola. Their intersection points are four points. Wait, but hyperbola and circle can intersect at up to four points. So yes, four points.

But the problem says they intersect at four points. So I need to find those four points, compute their distances to (-3, 2), and sum them up.

But computing four points and then calculating distance to (-3, 2) might be tedious. Maybe there's a smarter way?

Alternatively, maybe there is some geometric interpretation. Let me think.

First, note that (-3, 2) is the point... Let me check: The center of the first circle is at (-3, 12). So (-3, 2) is directly below the center of the first circle. The distance between (-3, 2) and (-3, 12) is 10 units. Since the radius of the first circle is 9, so (-3, 2) is inside the first circle because 10 > 9? Wait, distance from (-3, 2) to center (-3, 12) is 10 units. But radius is 9, so (-3, 2) is outside the first circle? Wait, distance from center to the point is 10, which is greater than radius 9, so yes, it's outside. So the point (-3, 2) is outside the first circle. However, since the second equation is a hyperbola, which is different.

But maybe there is a way to relate the distances?

Alternatively, maybe the problem can be approached by solving the equations and then computing the distances. Let me try to find the intersection points.

We have two equations:

1. $ (x + 3)^2 + (y - 12)^2 = 81 $ (Circle C1)
2. $ (y - 8)^2 - (x + 3)^2 = 9 $ (Hyperbola H)

So perhaps we can solve these two equations simultaneously. Let me denote:

From equation 1: Let me expand equation 1 again to see if I can use substitution with equation 2. Wait, maybe not. Let me note that equation 2 can be rearranged as (y - 8)^2 = (x + 3)^2 + 9.

So let me substitute this into equation 1. Equation 1 is:

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

But (y - 12) can be expressed in terms of (y - 8). Let me compute (y - 12) = (y - 8) - 4. So:

$ (x + 3)^2 + [(y - 8) - 4]^2 = 81 $

Which is:

$ (x + 3)^2 + (y - 8)^2 - 8(y - 8) + 16 = 81 $

But from equation 2, we know that (y - 8)^2 = (x + 3)^2 + 9. Let me substitute that here.

So replacing (y - 8)^2:

$ (x + 3)^2 + [(x + 3)^2 + 9] - 8(y - 8) + 16 = 81 $

Combine terms:

First, expand:

$ (x + 3)^2 + (x + 3)^2 + 9 - 8(y - 8) + 16 = 81 $

Combine like terms:

2*(x + 3)^2 + 25 - 8(y - 8) = 81

Subtract 25 from both sides:

2*(x + 3)^2 - 8(y - 8) = 56

Divide both sides by 2:

(x + 3)^2 - 4(y - 8) = 28

Hmm, not sure if helpful yet. Let me check if I can express y in terms of x or vice versa. Alternatively, maybe use equation 2 to express (x + 3)^2.

From equation 2: (y - 8)^2 - (x + 3)^2 = 9 => (x + 3)^2 = (y - 8)^2 - 9. Let me plug that into the previous equation.

So the equation above was:

(x + 3)^2 - 4(y - 8) = 28. Substitute (x + 3)^2:

(y - 8)^2 - 9 - 4(y - 8) = 28

Let me denote z = y - 8. Then equation becomes:

z^2 - 9 - 4z = 28 => z^2 - 4z - 37 = 0

Solving quadratic equation: z = [4 ± sqrt(16 + 148)] / 2 = [4 ± sqrt(164)] / 2 = [4 ± 2*sqrt(41)] / 2 = 2 ± sqrt(41)

Therefore, z = 2 + sqrt(41) or z = 2 - sqrt(41). Since z = y - 8, then y = 8 + z = 8 + 2 ± sqrt(41) = 10 ± sqrt(41)

So y = 10 + sqrt(41) and y = 10 - sqrt(41). Now, let's find corresponding x values.

From equation 2: (y - 8)^2 - (x + 3)^2 = 9 => (x + 3)^2 = (y - 8)^2 - 9

Since we have y, let's compute (y - 8)^2 for each case.

For y = 10 + sqrt(41):

(y - 8) = 2 + sqrt(41), so (y - 8)^2 = (2 + sqrt(41))^2 = 4 + 4*sqrt(41) + 41 = 45 + 4*sqrt(41)

Then, (x + 3)^2 = 45 + 4*sqrt(41) - 9 = 36 + 4*sqrt(41)

Similarly, for y = 10 - sqrt(41):

(y - 8) = 2 - sqrt(41), so (y - 8)^2 = (2 - sqrt(41))^2 = 4 - 4*sqrt(41) + 41 = 45 - 4*sqrt(41)

Then, (x + 3)^2 = 45 - 4*sqrt(41) - 9 = 36 - 4*sqrt(41)

So for each y value, we have two x values.

So for y = 10 + sqrt(41):

(x + 3)^2 = 36 + 4*sqrt(41). Therefore, x + 3 = ± sqrt(36 + 4*sqrt(41)), so x = -3 ± sqrt(36 + 4*sqrt(41))

Similarly, for y = 10 - sqrt(41):

(x + 3)^2 = 36 - 4*sqrt(41). Therefore, sqrt(36 - 4*sqrt(41))... Wait, is 36 - 4*sqrt(41) positive?

Compute sqrt(41) ≈ 6.403, so 4*sqrt(41) ≈ 25.612, so 36 - 25.612 ≈ 10.388, which is positive. Therefore, sqrt(36 - 4*sqrt(41)) is real, so x = -3 ± sqrt(36 - 4*sqrt(41))

Therefore, the four points of intersection are:

1. (-3 + sqrt(36 + 4*sqrt(41)), 10 + sqrt(41))
2. (-3 - sqrt(36 + 4*sqrt(41)), 10 + sqrt(41))
3. (-3 + sqrt(36 - 4*sqrt(41)), 10 - sqrt(41))
4. (-3 - sqrt(36 - 4*sqrt(41)), 10 - sqrt(41))

That seems complicated. Now, I need to compute the sum of distances from each of these four points to (-3, 2).

Let me denote the four points as P1, P2, P3, P4.

Let me note that each point has x-coordinate of the form -3 ± something, and y-coordinate of 10 ± sqrt(41). Let me think about the distance formula.

For a general point (x, y), distance to (-3, 2) is sqrt[(x + 3)^2 + (y - 2)^2]

For each point:

For P1: x = -3 + sqrt(36 + 4*sqrt(41)), y = 10 + sqrt(41)

So distance squared is [sqrt(36 + 4*sqrt(41))]^2 + [ (10 + sqrt(41) - 2) ]^2 = (36 + 4*sqrt(41)) + [8 + sqrt(41)]^2

Similarly for P2: x = -3 - sqrt(...), so [sqrt(...)]^2 is same as above, but since it's negative, when squared it's same. Therefore, distance squared for P2 is same as P1. Similarly for P3 and P4.

Therefore, the sum of distances squared would be 2*(distance squared of P1 + distance squared of P3). However, the problem asks for the sum of distances, not squared distances. Hmm. But maybe there's something with symmetry?

Alternatively, perhaps the distances have some relation. Let me think.

Alternatively, maybe we can compute the sum of distances by using properties of circles and hyperbola.

Wait, but the point (-3, 2) is interesting. Let me check. The center of the first circle is (-3, 12). The point (-3, 2) is directly below that center. Also, the equation of the first circle is (x + 3)^2 + (y - 12)^2 = 81. So, (-3, 2) is a point outside of this circle because the distance between them is 10, which is larger than the radius 9.

But let me think: If we consider the four points of intersection between the circle and hyperbola. Maybe they lie on some relation with respect to (-3, 2). Let me consider that.

Alternatively, maybe I can use parametrization or coordinates. Let me try to compute the distances.

First, for P1 and P2, their x-coordinates are symmetric around -3, and their y-coordinate is same. Similarly for P3 and P4.

Let me compute the distance for P1:

Distance from (-3, 2) to P1:

sqrt[ (x - (-3))^2 + (y - 2)^2 ] = sqrt[ (x + 3)^2 + (y - 2)^2 ]

But x + 3 is sqrt(36 + 4√41) or -sqrt(...), but squared it's same. So for P1 and P2, distance squared is:

(x + 3)^2 + (y - 2)^2 = (36 + 4√41) + (10 + √41 - 2)^2 = 36 + 4√41 + (8 + √41)^2

Similarly, compute (8 + √41)^2:

8^2 + 2*8*√41 + (√41)^2 = 64 + 16√41 + 41 = 105 + 16√41

Therefore, total distance squared for P1 and P2 combined:

2*(36 + 4√41 + 105 + 16√41) = 2*(141 + 20√41) ?

Wait, wait. Wait, no: For each of P1 and P2, their distance squared is 36 + 4√41 + 105 + 16√41? Wait no, that would be if P1 and P2 have same distance squared. Wait, let me redo:

For P1: distance squared is (x + 3)^2 + (y - 2)^2 = [sqrt(36 + 4√41)]^2 + [ (10 + √41 - 2) ]^2 = 36 + 4√41 + (8 + √41)^2. As above, which is 36 + 4√41 + 105 + 16√41 = 36 + 105 + 4√41 + 16√41 = 141 + 20√41.

Similarly for P2, same value. Therefore, sum of distances squared for P1 and P2 is 2*(141 + 20√41). Similarly for P3 and P4.

Let me compute for P3 and P4:

For P3, y = 10 - sqrt(41), so (y - 2) = 8 - sqrt(41). Then, (y - 2)^2 = (8 - sqrt(41))^2 = 64 - 16√41 + 41 = 105 - 16√41.

And (x + 3)^2 is 36 - 4√41. Therefore, distance squared for P3 is:

36 - 4√41 + 105 - 16√41 = 141 - 20√41. Similarly for P4. So total distance squared for P3 and P4 is 2*(141 - 20√41).

Therefore, total sum of squared distances is 2*(141 + 20√41) + 2*(141 - 20√41) = 2*141 + 2*20√41 + 2*141 - 2*20√41 = 4*141 = 564.

But the problem asks for the sum of distances, not the sum of squared distances. However, maybe there is a way to relate the sum of distances to the sum of squared distances? Not sure. Alternatively, maybe we can consider complex numbers or something else?

Alternatively, perhaps there is a geometric approach. Let me think again.

Wait, if I think about the sum of distances from four points to a fixed point. But unless there's some symmetry, it's difficult. But let me think about the four points.

Note that the first circle has center (-3, 12) and radius 9. The point (-3, 2) is along the vertical line x = -3, 10 units below the center.

Also, the hyperbola equation is (y - 8)^2 - (x + 3)^2 = 9. Which is a hyperbola centered at (-3, 8). So the center of the hyperbola is at (-3, 8). The circle is centered at (-3, 12), which is 4 units above the hyperbola's center.

But how does this help? Maybe not directly.

Alternatively, maybe think of the four points as lying on both the circle and the hyperbola. Let me check if these points can be related to (-3, 2).

Alternatively, maybe use complex numbers or parametric equations. Alternatively, think about inversion? Not sure.

Alternatively, since the problem is asking for the sum of distances, and we have expressions for the squared distances. Maybe we can compute the sum of square roots?

But that seems complicated. However, note that if we have sum of distances, and we know sum of squared distances, but unless we have more info, it's not helpful. But maybe there is a way.

Alternatively, perhaps the four points lie on a circle with diameter between (-3, 2) and something? Not sure.

Alternatively, think about Vieta's formula. Wait, maybe if we consider the four points on both curves, and relate to coordinates. But maybe not straightforward.

Wait, let me go back. Earlier, I calculated the sum of squared distances to (-3, 2) as 564. But maybe there's a way to compute the sum of distances?

Alternatively, maybe note that the four points lie on the circle C1. Let me check.

Yes, all four points lie on the circle C1. So, for each point (x, y) on C1, we have (x + 3)^2 + (y - 12)^2 = 81.

Let me denote P = (-3, 2). Let me compute the distance from P to any point on C1.

But the sum of distances from four points on the circle to P... Not sure. But maybe there's some relation. Wait, but the four points are not arbitrary on the circle. They also lie on the hyperbola.

Alternatively, maybe parametrize the circle and find points also on the hyperbola? But that seems similar to what I did before.

Wait, here's an idea. Since the points lie on both the circle and hyperbola, maybe I can find a relation between (x + 3)^2 and (y - 8)^2 from the hyperbola equation.

From hyperbola equation: (y - 8)^2 - (x + 3)^2 = 9 => (x + 3)^2 = (y - 8)^2 - 9.

From circle equation: (x + 3)^2 + (y - 12)^2 = 81. Substitute (x + 3)^2:

(y - 8)^2 - 9 + (y - 12)^2 = 81 => (y - 8)^2 + (y - 12)^2 = 90.

Expanding both squares:

(y² - 16y + 64) + (y² - 24y + 144) = 90

Combine terms:

2y² - 40y + 208 = 90 => 2y² - 40y + 118 = 0 => Divide by 2: y² - 20y + 59 = 0

Wait, earlier when solving for y, we had y = 10 ± sqrt(41). Let me check if that satisfies y² - 20y + 59 = 0.

Let me compute for y = 10 + sqrt(41):

y² = (10 + sqrt(41))² = 100 + 20*sqrt(41) + 41 = 141 + 20*sqrt(41)

Then y² - 20y + 59 = (141 + 20√41) - 20*(10 + √41) + 59 = 141 + 20√41 - 200 - 20√41 + 59 = (141 - 200 + 59) + (20√41 - 20√41) = 0. Correct. Similarly for the other root.

So, this gives us y² - 20y + 59 = 0.

So for each of the four points, their y-coordinates satisfy this quadratic. Also, the x-coordinates can be found using the earlier equations. But how does this help?

Alternatively, think about the distances from (-3, 2) to each point. Let me denote D as the distance from (-3, 2) to a point (x, y). Then, D² = (x + 3)² + (y - 2)².

But from the circle equation, (x + 3)² = 81 - (y - 12)². Therefore, D² = 81 - (y - 12)² + (y - 2)².

Compute that:

D² = 81 - [ (y² - 24y + 144) - (y² - 4y + 4) ]? Wait, wait:

Wait, (y - 12)^2 is y² - 24y + 144, and (y - 2)^2 is y² - 4y + 4. Therefore:

D² = 81 - [ (y² - 24y + 144) ] + (y² - 4y + 4) = 81 - y² + 24y - 144 + y² - 4y + 4.

Simplify terms:

- y² + y² cancels.

Then 81 - 144 + 4 + 24y - 4y = (-59) + 20y. So D² = 20y - 59.

Wow! Therefore, for each of the four points, D² = 20y - 59. Therefore, D = sqrt(20y - 59).

But since we have four points, each with y-coordinate being 10 ± sqrt(41), let me check that.

For the two values of y:

First case: y = 10 + sqrt(41). Then D² = 20*(10 + sqrt(41)) - 59 = 200 + 20*sqrt(41) - 59 = 141 + 20*sqrt(41).

Which matches with what I had earlier. Similarly, for y = 10 - sqrt(41): D² = 20*(10 - sqrt(41)) - 59 = 200 - 20*sqrt(41) - 59 = 141 - 20*sqrt(41).

So, for each of the four points, their distances squared are 141 ± 20√41 (each occurring twice).

But the problem asks for the sum of distances, not squared distances. However, perhaps I can find the sum of sqrt(20y - 59) for each of the four points.

But since two of them have y = 10 + sqrt(41) and two have y = 10 - sqrt(41), then sum of distances is 2*sqrt(141 + 20√41) + 2*sqrt(141 - 20√41).

So, sum = 2[ sqrt(141 + 20√41) + sqrt(141 - 20√41) ]

Hmm, need to compute this expression. Let me denote A = sqrt(141 + 20√41) + sqrt(141 - 20√41). Then sum is 2A. Let me compute A.

Let me compute A²:

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

= (sqrt(a) + sqrt(b))² where a = 141 + 20√41, b = 141 - 20√41

= a + b + 2*sqrt(ab)

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

Compute ab = (141 + 20√41)(141 - 20√41) = 141² - (20√41)²

Calculate:

141²: Well, 140² = 19600, 140² + 2*140 + 1 = 19600 + 280 + 1 = 19881.

(20√41)² = 400 * 41 = 16400

So ab = 19881 - 16400 = 3481.

So sqrt(ab) = sqrt(3481). Let me check sqrt(3481).

Well, 59² = 3481 because 60²=3600, so 59² = 3481. Yes. Therefore, sqrt(ab) = 59.

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

Therefore, A = sqrt(400) = 20.

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

Wow! So the answer is 40.

But let me just check my steps again to make sure I didn't make a mistake.

First, I found that D² = 20y - 59 for each point. Then, since y for two points is 10 + sqrt(41) and two points 10 - sqrt(41), so each of their D squared is 20*(10 ± sqrt(41)) - 59. Then, the sum is 2*sqrt(20*(10 + sqrt(41)) -59) + 2*sqrt(...).

Then, I computed that expression and found that A² = 400, so A = 20, so total sum is 40. Seems correct.

Alternatively, maybe there's a smarter way to see that the sum is 40.

Alternatively, note that when I found D² = 20y - 59, then for each point on the circle, since two points have y = 10 + sqrt(41) and two have y = 10 - sqrt(41), so total sum of D is 2*(sqrt(20*(10 + sqrt(41)) -59) + sqrt(20*(10 - sqrt(41)) -59)) )

But then, simplifying inside the square roots:

20*(10 + sqrt(41)) -59 = 200 + 20√41 -59 = 141 + 20√41

Similarly for the other term. Then, we found that sqrt(141 + 20√41) + sqrt(141 - 20√41) = 20.

Therefore, total sum is 2*20 = 40.

Therefore, the answer is 40.

Alternatively, maybe there is a way to see that the sum of sqrt(141 + 20√41) + sqrt(141 - 20√41) equals 20? Let me check with numbers.

Let me compute sqrt(141 + 20√41) + sqrt(141 - 20√41). Suppose this equals 20. Then, let me compute sqrt(141 + 20√41). Let me approximate values.

First, sqrt(41) ≈ 6.403. Then 20*sqrt(41) ≈ 128.06. Therefore, 141 + 128.06 ≈ 269.06. sqrt(269.06) ≈ 16.4.

Similarly, 141 - 128.06 ≈ 12.94. sqrt(12.94) ≈ 3.597. Then 16.4 + 3.597 ≈ 19.997 ≈ 20. So yes, it checks out numerically.

Therefore, the answer is 40.

Therefore, I think that the answer is 40.

**Final Answer**
\boxed{40}

To solve the problem, we are given two conic equations:

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

The goal is to compute the **sum of the distances** from the four intersection points of these conics to the point $(-3, 2)$.

---

### Step 1: Recognize the Shapes of the Curves

- The first equation is a **circle**, and after completing the square:
  $$
  (x + 3)^2 + (y - 12)^2 = 81
  $$
  This is a circle with **center at (-3, 12)** and **radius 9**.

- The second equation is a **hyperbola**, and after rearranging:
  $$
  (y - 8)^2 - (x + 3)^2 = 9
  $$
  This is a vertical hyperbola centered at $(-3, 8)$, opening up and down.

---

### Step 2: Find Coordinates of the Intersection Points

By solving both equations simultaneously, we find that the four points of intersection lie at:

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

This gives us four distinct points of intersection.

---

### Step 3: Compute the Sum of Distances

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

Let’s define the distance from a point $ (x, y) $ to $(-3, 2)$ as:
$$
D = \sqrt{(x + 3)^2 + (y - 2)^2}
$$

We simplify using the fact that all four points lie on the circle centered at $(-3, 12)$, so we can substitute the circle's equation:
$$
(x + 3)^2 + (y - 12)^2 = 81 \Rightarrow (x + 3)^2 = 81 - (y - 12)^2
$$

Then the square of the distance becomes:
$$
D^2 = (x + 3)^2 + (y - 2)^2 = 81 - (y - 12)^2 + (y - 2)^2
$$

Simplifying:
$$
D^2 = 81 - [(y - 12)^2 - (y - 2)^2]
$$

Expanding both squared terms:
$$
(y - 12)^2 = y^2 - 24y + 144,\quad (y - 2)^2 = y^2 - 4y + 4
$$
$$
D^2 = 81 - [y^2 - 24y + 144 - (y^2 - 4y + 4)] = 81 - [-20y + 140] = 81 + 20y - 140 = 20y - 59
$$

Therefore, the distance from each point to (-3, 2) is:
$$
D = \sqrt{20y - 59}
$$

---

### Step 4: Use Symmetry and Roots

From earlier steps, we found that the **y-coordinates** of the four points satisfy:
$$
y^2 - 20y + 59 = 0
$$
Thus, the two distinct y-values are:
$$
y = 10 \pm \sqrt{41}
$$

So the four distance contributions are:
- Two points with $ y = 10 + \sqrt{41} \Rightarrow D = \sqrt{20(10 + \sqrt{41}) - 59} = \sqrt{141 + 20\sqrt{41}} $
- Two points with $ y = 10 - \sqrt{41} \Rightarrow D = \sqrt{20(10 - \sqrt{41}) - 59} = \sqrt{141 - 20\sqrt{41}} $

Therefore, the **sum of distances** is:
$$
\text{Sum} = 2\left( \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}} \right)
$$

Let us denote:
$$
A = \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}}
$$

We square $A$:
$$
A^2 = \left( \sqrt{141 + 20\sqrt{41}} + \sqrt{141 - 20\sqrt{41}} \right)^2
$$
$$
= (141 + 20\sqrt{41}) + (141 - 20\sqrt{41}) + 2\sqrt{(141 + 20\sqrt{41})(141 - 20\sqrt{41})}
$$

Simplify:
- $141 + 20\sqrt{41} + 141 - 20\sqrt{41} = 282$
- $(141 + 20\sqrt{41})(141 - 20\sqrt{41}) = 141^2 - (20\sqrt{41})^2 = 19881 - 16400 = 3481$

So:
$$
A^2 = 282 + 2\sqrt{3481} = 282 + 2 \cdot 59 = 282 + 118 = 400
\Rightarrow A = \sqrt{400} = 20
$$

Hence, the total distance is:
$$
2A = 2 \cdot 20 = \boxed{40}
$$

---

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