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 given equations are:

1. \(x^2 + y^2 + 6x - 24y + 72 = 0\)
2. \(x^2 - y^2 + 6x + 16y - 46 = 0\)

Hmm, the first equation looks like a circle equation, but the second one has a \(y^2\) term with a negative coefficient. Wait, that means the second equation isn't a circle but a hyperbola. So, we have a circle and a hyperbola intersecting at four points. The problem is asking for the sum of the distances from these four points to the point (-3,2). 

Alright, so maybe I should first find the points of intersection between the circle and the hyperbola. Once I have those four points, I can compute the distance from each to (-3,2) and sum them up. Let me try that approach.

Starting with the two equations:

Equation 1: \(x^2 + y^2 + 6x - 24y + 72 = 0\)

Equation 2: \(x^2 - y^2 + 6x + 16y - 46 = 0\)

I can try subtracting one equation from the other to eliminate \(x^2\) terms. Let me subtract Equation 2 from Equation 1:

Equation 1 - Equation 2:

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

Simplify term by term:

x^2 - x^2 = 0

y^2 - (-y^2) = 2y^2

6x - 6x = 0

-24y - 16y = -40y

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

So the result is:

2y^2 - 40y + 118 = 0

Divide both sides by 2 to simplify:

y^2 - 20y + 59 = 0

Hmm, solving this quadratic for y. Let's use the quadratic formula:

y = [20 ± sqrt(400 - 236)] / 2

sqrt(400 - 236) = sqrt(164) = 2*sqrt(41)

So y = [20 ± 2sqrt(41)] / 2 = 10 ± sqrt(41)

So the y-coordinates of the intersection points are y = 10 + sqrt(41) and y = 10 - sqrt(41). 

Now, to find the corresponding x-coordinates, I can plug these y-values back into one of the original equations. Let's pick Equation 2 since it's a hyperbola and might be simpler. Equation 2 is:

x^2 - y^2 + 6x + 16y - 46 = 0

Let me rearrange this equation to solve for x^2:

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

Completing the square for x:

x^2 + 6x + 9 = y^2 - 16y + 46 + 9

So (x + 3)^2 = y^2 - 16y + 55

Therefore, x + 3 = ±sqrt(y^2 - 16y + 55)

Thus, x = -3 ± sqrt(y^2 - 16y + 55)

Alternatively, since we have specific y-values, we can plug them in. Let's compute for each y:

First, for y = 10 + sqrt(41):

Compute y^2:

(10 + sqrt(41))^2 = 100 + 20sqrt(41) + 41 = 141 + 20sqrt(41)

Then, -16y = -16*(10 + sqrt(41)) = -160 -16sqrt(41)

So y^2 -16y +55 = (141 + 20sqrt(41)) + (-160 -16sqrt(41)) +55

Compute constants: 141 -160 +55 = 36

sqrt terms: 20sqrt(41) -16sqrt(41) = 4sqrt(41)

So y^2 -16y +55 = 36 +4sqrt(41)

Therefore, sqrt(y^2 -16y +55) = sqrt(36 +4sqrt(41))

Hmm, this seems complicated. Maybe there's a better way. Alternatively, since we have both equations, perhaps we can substitute y from the first equation into the second.

Wait, maybe I can use Equation 1 to express x^2 + y^2 in terms of x and y, then substitute into Equation 2.

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

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

From Equation 1, x^2 + y^2 = -6x +24y -72

Let me substitute x^2 + y^2 into Equation 2. Equation 2 can be written as:

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

But from Equation 1, x^2 + y^2 = -6x +24y -72. Substitute that into Equation 2:

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

Simplify term by term:

-6x +6x = 0

24y +16y =40y

-72 -46 = -118

So the equation becomes:

-2y^2 +40y -118 =0

Multiply both sides by -1:

2y^2 -40y +118 =0

Which is the same as before. So that's consistent. So we have the same quadratic, leading to y =10 ±sqrt(41). So that seems correct.

So now, to find x, perhaps substitute back into Equation 1 or 2. Let's try Equation 2 again.

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

We can write this as x^2 +6x = y^2 -16y +46

Completing the square for x:

x^2 +6x +9 = y^2 -16y +55

So (x +3)^2 = y^2 -16y +55

Therefore, x = -3 ± sqrt(y^2 -16y +55)

So for each y, we can compute sqrt(y^2 -16y +55) and then get x.

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

Compute y^2: (10 + sqrt(41))^2 =100 +20sqrt(41) +41=141 +20sqrt(41)

Then y^2 -16y =141 +20sqrt(41) -16*(10 + sqrt(41))=141 +20sqrt(41) -160 -16sqrt(41)= (141-160)+(20sqrt(41)-16sqrt(41))= -19 +4sqrt(41)

Then y^2 -16y +55= -19 +4sqrt(41) +55=36 +4sqrt(41)

So sqrt(y^2 -16y +55)=sqrt(36 +4sqrt(41))

Hmm, this is still complicated. Let me rationalize or see if this can be simplified. Let me check if sqrt(36 +4sqrt(41)) can be expressed as sqrt(a) + sqrt(b). Let's suppose sqrt(36 +4sqrt(41))=sqrt(a)+sqrt(b). Then squaring both sides:

36 +4sqrt(41)=a + b + 2sqrt(ab)

Comparing terms:

a + b =36

2sqrt(ab)=4sqrt(41) => sqrt(ab)=2sqrt(41) => ab=4*41=164

So we have a + b =36 and ab=164. We need to find a and b such that they satisfy these equations.

Let me solve the quadratic equation: x^2 -36x +164=0

Discriminant: 1296 -656=640

sqrt(640)=8sqrt(10)

So a=(36 ±8sqrt(10))/2=18 ±4sqrt(10)

Hmm, not integers. So maybe sqrt(36 +4sqrt(41)) cannot be simplified into sqrt(a)+sqrt(b). Therefore, perhaps it's better to just leave it as sqrt(36 +4sqrt(41)).

Alternatively, maybe we can compute the numerical value. Let's compute sqrt(41) first: sqrt(41)≈6.4031

Then 36 +4*6.4031≈36 +25.6124≈61.6124

sqrt(61.6124)≈7.85

So approximately, sqrt(36 +4sqrt(41))≈7.85

Therefore, x≈-3 ±7.85

So x≈4.85 or x≈-10.85

But since we need exact values, maybe there's another way. Let me check if the expressions can be simplified.

Alternatively, maybe we can use the fact that the sum of distances from the four intersection points to (-3,2) can be found using properties of circles or hyperbolas. Wait, the point (-3,2) is given. Let me check if this point has any special relation to the equations.

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

Let me complete the squares for this equation.

x^2 +6x = (x+3)^2 -9

y^2 -24y = (y-12)^2 -144

So the equation becomes:

(x+3)^2 -9 + (y-12)^2 -144 +72=0

Simplify constants: -9 -144 +72= -81

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

Therefore, Equation 1 is a circle with center at (-3,12) and radius 9.

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

Earlier, we tried to rearrange it, but maybe let's also complete the squares for this equation.

x^2 +6x = (x+3)^2 -9

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

So substituting back into Equation 2:

(x+3)^2 -9 - (y-8)^2 +64 -46=0

Simplify constants: -9 +64 -46=9

So (x+3)^2 - (y-8)^2 +9=0

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

Which is a hyperbola centered at (-3,8), opening upwards and downwards, since the y-term is negative. The standard form is ((x-h)^2)/a^2 - ((y-k)^2)/b^2 = -1, which is equivalent to ((y - k)^2)/b^2 - ((x - h)^2)/a^2 =1. So this hyperbola opens along the y-axis.

Therefore, the two curves are a circle centered at (-3,12) with radius 9 and a hyperbola centered at (-3,8). The point (-3,2) is the center of the circle shifted down by 10 units. Let me see if that's on the hyperbola or not.

Wait, maybe there's a reflection property or something here. The sum of distances from four points to a fixed point. But since it's four points, maybe there's a way to relate this to the properties of the circle and hyperbola.

Alternatively, perhaps the four intersection points lie on both curves, so maybe we can use some geometric properties. But I'm not sure. Maybe it's better to find the coordinates of the four intersection points and then compute the distances.

But since solving for y gave us y=10 ±sqrt(41), and then x=-3 ±sqrt(36 +4sqrt(41)), which seems complicated, perhaps there's a smarter way.

Wait, let me think. The sum of distances from four points to a fixed point. If the four points lie on a circle, then the sum of distances from them to a fixed point might relate to the power of the point with respect to the circle. But here, the four points are on both the circle and the hyperbola, not necessarily on a circle. But maybe the sum can be found using some algebraic properties.

Alternatively, maybe using complex numbers or vectors. Hmm, not sure.

Alternatively, perhaps the sum can be expressed in terms of the coordinates of the intersection points. Let me denote the four points as (x1, y1), (x2, y2), (x3, y3), (x4, y4). Then the sum of distances is sqrt((x1 +3)^2 + (y1 -2)^2) + sqrt((x2 +3)^2 + (y2 -2)^2) + sqrt((x3 +3)^2 + (y3 -2)^2) + sqrt((x4 +3)^2 + (y4 -2)^2). But this seems too complicated.

Alternatively, maybe there is a way to parameterize the intersection points. Since we have two equations, maybe we can find a parametric equation for the intersection.

But since the equations are a circle and a hyperbola, solving them simultaneously leads to quartic equations, which might have four solutions. But in our case, we already found the y-coordinates, which are y=10 + sqrt(41) and y=10 - sqrt(41). For each y, there are two x-values, so four points in total. 

But computing each x and y exactly is messy. Maybe there's a trick here. Let me check if the sum of the distances can be simplified without knowing the exact coordinates.

Alternatively, maybe note that the sum of distances from four points on a circle to a fixed point inside the circle can be related to the power of the point or something. Wait, but the fixed point (-3,2) is not necessarily inside the circle. Let me check the distance from (-3,2) to the center of the circle (-3,12). The distance is |12 -2|=10. The radius of the circle is 9. So the point (-3,2) is outside the circle, since the distance from the center is 10 >9. So the power of the point with respect to the circle is (distance)^2 - radius^2 =100 -81=19. The power of a point outside a circle is equal to the square of the tangent length from the point to the circle. But how does that relate to the sum of distances?

Alternatively, the sum of distances from four points on a circle to an external point. But I don't recall a formula for that. Maybe if the points are symmetric with respect to the center, but in this case, the hyperbola's center is at (-3,8), which is different from the circle's center at (-3,12). So the four intersection points are symmetric in some way?

Wait, looking back at the y-coordinates: 10 + sqrt(41) and 10 - sqrt(41). These are symmetric around 10. Similarly, the hyperbola is centered at (-3,8), so maybe the x-coordinates are symmetric around -3? Let me check.

From Equation 2, when we solved for x, we had x = -3 ± sqrt(y^2 -16y +55). So for each y, the x-coordinates are symmetric around -3. Therefore, for each pair of points with the same y, their x-coordinates are -3 + sqrt(...) and -3 - sqrt(...). So the four points are symmetric with respect to the vertical line x = -3.

Therefore, when we compute the distances from these points to (-3,2), which is also on the vertical line x = -3, the horizontal distance (x-coordinate difference) will be the same for points symmetric around x=-3. 

Therefore, for each pair of points symmetric around x=-3, their horizontal distance to (-3,2) is the same. Let me elaborate.

Suppose one point is (-3 + a, y1), and the other is (-3 - a, y1). Then the distance from (-3 + a, y1) to (-3,2) is sqrt(a^2 + (y1 -2)^2), and the distance from (-3 -a, y1) to (-3,2) is sqrt(a^2 + (y1 -2)^2). So both distances are equal. Therefore, each pair contributes twice the same distance. Similarly, for the other pair of points with the same y-coordinate, they would also contribute twice the same distance.

Therefore, the total sum of distances would be 2*(sqrt(a^2 + (y1 -2)^2) + sqrt(a^2 + (y2 -2)^2)), where a is the sqrt(...) term, and y1 and y2 are the two distinct y-coordinates.

But wait, since we have four points, two pairs, each pair has the same a but different y1 and y2. So maybe more precisely, the sum is 2*[sqrt(a1^2 + (y1 -2)^2) + sqrt(a2^2 + (y2 -2)^2)], where a1 and a2 are sqrt(y1^2 -16y1 +55) and sqrt(y2^2 -16y2 +55) respectively.

But this still requires computing a1 and a2, which are sqrt(36 +4sqrt(41)) and sqrt(36 -4sqrt(41)), since y1=10 +sqrt(41), y2=10 -sqrt(41).

Wait, let me compute a1 and a2:

For y1 =10 +sqrt(41):

a1 = sqrt(y1^2 -16y1 +55) = sqrt(36 +4sqrt(41)) as before.

For y2=10 -sqrt(41):

a2 = sqrt(y2^2 -16y2 +55) = sqrt((10 -sqrt(41))^2 -16*(10 -sqrt(41)) +55)

Compute this:

(10 -sqrt(41))^2 =100 -20sqrt(41) +41=141 -20sqrt(41)

-16*(10 -sqrt(41))= -160 +16sqrt(41)

So y2^2 -16y2 +55 = (141 -20sqrt(41)) + (-160 +16sqrt(41)) +55

Constants:141 -160 +55=36

sqrt terms: -20sqrt(41) +16sqrt(41)= -4sqrt(41)

So y2^2 -16y2 +55=36 -4sqrt(41)

Therefore, a2= sqrt(36 -4sqrt(41))

Therefore, the sum of distances is 2*(sqrt(36 +4sqrt(41) + (10 +sqrt(41) -2)^2) + sqrt(36 -4sqrt(41) + (10 -sqrt(41) -2)^2)).

Wait, hold on. Let me re-examine. The distance from each point (-3 + a1, y1) to (-3,2) is sqrt(a1^2 + (y1 -2)^2). Similarly for the other points. Therefore, the total sum is 2*[sqrt(a1^2 + (y1 -2)^2) + sqrt(a2^2 + (y2 -2)^2)].

But let's compute (y1 -2)^2 and (y2 -2)^2.

y1=10 +sqrt(41), so y1 -2=8 +sqrt(41). Therefore, (y1 -2)^2=(8 +sqrt(41))^2=64 +16sqrt(41)+41=105 +16sqrt(41)

Similarly, y2=10 -sqrt(41), so y2 -2=8 -sqrt(41). Therefore, (y2 -2)^2=(8 -sqrt(41))^2=64 -16sqrt(41)+41=105 -16sqrt(41)

Therefore, the sum becomes:

2*[sqrt(a1^2 +105 +16sqrt(41)) + sqrt(a2^2 +105 -16sqrt(41))]

But a1^2=36 +4sqrt(41), a2^2=36 -4sqrt(41)

So:

sqrt((36 +4sqrt(41)) +105 +16sqrt(41)) = sqrt(141 +20sqrt(41))

sqrt((36 -4sqrt(41)) +105 -16sqrt(41)) = sqrt(141 -20sqrt(41))

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

Now, let me compute sqrt(141 +20sqrt(41)) + sqrt(141 -20sqrt(41)). Let me denote this as S.

Compute S^2 = [sqrt(141 +20sqrt(41)) + sqrt(141 -20sqrt(41))]^2

= (141 +20sqrt(41)) + (141 -20sqrt(41)) + 2*sqrt{(141 +20sqrt(41))(141 -20sqrt(41))}

Simplify:

141 +141 =282

20sqrt(41) -20sqrt(41) =0

The cross term: 2*sqrt{(141)^2 - (20sqrt(41))^2} =2*sqrt{19881 - 400*41} =2*sqrt{19881 -16400}=2*sqrt{3481}=2*59=118

Therefore, S^2=282 +118=400

Therefore, S=20

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

Wait, that's a nice result! So the total sum is 40. That seems surprisingly clean.

So, to recap:

1. Found intersection points have y-coordinates 10 ±sqrt(41).

2. For each y, x-coordinates are -3 ±sqrt(36 ±4sqrt(41)).

3. For each pair of points with the same y, their distances to (-3,2) are equal, so the total sum is 2*(distance1 + distance2).

4. Calculated distance1 and distance2 by recognizing that the sum S = sqrt(141 +20sqrt(41)) + sqrt(141 -20sqrt(41)) equals 20, hence total sum is 40.

Therefore, the answer is 40.

But let me verify once more.

We had S^2 = [sqrt(a) + sqrt(b)]^2 = a + b + 2sqrt(ab). Here, a=141 +20sqrt(41), b=141 -20sqrt(41). Then a + b=282. ab=(141)^2 - (20sqrt(41))^2=19881 -16400=3481. sqrt(ab)=sqrt(3481)=59. Therefore, S^2=282 +2*59=282 +118=400, so S=20. Then total sum is 2*20=40. Yes, that's correct.

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

**Final Answer**
\boxed{40}
Therefore, the sum of the distances from these four points to the point \((-3,2)\) is $\boxed{40}$.
</think>