<Question>
The graphs of $x^2 + y^2 + 6x - 24y + 72 = 0$ and $x^2 - y^2 + 6x + 16y - 46 = 0$ intersect at four points.  Compute the sum of the distances from these four points to the point $(-3,2).$
</Question>

<think>
<Plan_1>
Okay, so I need to find the sum of the distances from the four intersection points of these two graphs to the point (-3, 2). Hmm, let me start by understanding what the problem is asking. There are two conic sections given by their equations, and they intersect at four points. For each of these four points, I have to calculate the distance to (-3, 2) and then sum all those distances. The challenge here is that solving for the four intersection points might be a bit involved, but maybe there's a smarter way to do this without having to find each point individually.

First, let me write down the equations again to make sure I have them right:

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

These look like two conic sections. The first one, with all the \( x^2 \) and \( y^2 \) terms, might be a circle or an ellipse. The second one has an \( x^2 \) and a \( -y^2 \), which suggests it's a hyperbola. Let me check if the first equation is indeed a circle.

<prompt_1.1> Identify the type of conic section for each equation and plan to find their intersection points. </prompt_1.1>

Hmm, maybe adding the equations isn't helpful. Let me subtract them instead.

<prompt_1.2> Consider alternative methods to find the intersection points. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
For the first equation, if I rearrange terms:

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

Maybe I can complete the squares for both x and y terms.

For the x terms: \( x^2 + 6x \). Take half of 6, which is 3, square it to get 9. So, \( x^2 + 6x = (x + 3)^2 - 9 \).

For the y terms: \( y^2 - 24y \). Half of -24 is -12, squaring gives 144. So, \( y^2 - 24y = (y - 12)^2 - 144 \).

Substituting back into the equation:

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

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

\( (x + 3)^2 + (y - 12)^2 - 81 = 0 \)

Which becomes:

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

Ah, that's a circle with center at (-3, 12) and radius 9.
</execution_1.1>

<execution_1.2>
Wait, if I subtract the second equation from the first, maybe that will eliminate some terms. Let's try that:

First equation minus the second equation:

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

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:

\( y^2 - 20y + 59 = 0 \)

Hmm, this is a quadratic in y. Let me check the discriminant to see if there are real solutions. Discriminant D = 400 - 236 = 164. Since D is positive, there are two real solutions for y. So, the two curves intersect at two points with the same x-coordinate? Wait, but the problem states there are four intersection points. Maybe each y corresponds to two x's? Or maybe I made a mistake here.

Wait, let's think again. Subtracting the two equations eliminated the \( x^2 \) and \( 6x \) terms, leaving us with a quadratic in y. That suggests that for each x, there are two possible y's? But the discriminant is positive, so two y's. But how does that lead to four intersection points? Maybe each y corresponds to two x's?
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
The second equation is a hyperbola. Let's confirm:

<prompt_2.1> Identify the type of conic section for the second equation. </prompt_2.1>
-----

Based on execution_1.2:
Alternatively, maybe solving for x in terms of y from one equation and substituting into the other.

Alternatively, maybe parametrize one equation and substitute into the other. But perhaps there's a smarter way.

Wait, the problem is to compute the sum of distances from the four intersection points to (-3, 2). Maybe there's a way to use properties of conic sections or the radical axis?

<prompt_2.2> Explore alternative methods to find the intersection points. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
\( x^2 - y^2 + 6x + 16y - 46 = 0 \)

Let me try to rearrange and complete the squares here as well.

Group x terms and y terms:

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

For x: \( x^2 + 6x = (x + 3)^2 - 9 \)

For y: \( -y^2 + 16y = -(y^2 - 16y) \). Completing the square inside the parentheses: y^2 -16y = (y - 8)^2 - 64. So, -(y^2 -16y) = - ( (y - 8)^2 - 64 ) = - (y - 8)^2 + 64

Therefore, substituting back:

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

Simplify constants: -9 + 64 = 55. So:

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

Which simplifies to:

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

Divide both sides by -9 to get it into standard form:

\( \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 \)

So, this is a hyperbola centered at (-3, 8), opening upwards and downwards, with a transverse axis of length 6 (since a^2 = 9, so a = 3).
</execution_2.1>

<execution_2.2>
Wait, subtracting the two equations gives the radical axis of the two conic sections, which is the equation 2y² - 40y + 118 = 0, which simplifies to y² - 20y + 59 = 0. So the radical axis is two horizontal lines? Wait, solving for y, we get y = [20 ± sqrt(400 - 236)] / 2 = [20 ± sqrt(164)] / 2 = [20 ± 2*sqrt(41)] / 2 = 10 ± sqrt(41). So the two y-values where the intersection points lie are y = 10 + sqrt(41) and y = 10 - sqrt(41). Then, substituting these back into one of the original equations to find the corresponding x-values.

But since there are two y-values and each y would correspond to two x-values (since the first equation is a circle, which is a quadratic in x), that would give four points total. So, to find the four points, I need to solve for x when y is 10 + sqrt(41) and 10 - sqrt(41). Then, for each of these y's, plug into one of the equations and solve for x. Then, once I have all four (x, y) points, compute their distances to (-3, 2) and sum them up.

But this seems tedious, especially with the square roots. Maybe there's a smarter way. Let me check if the point (-3, 2) has any special relationship with the two conic sections.

Wait, the first circle is centered at (-3, 12) with radius 9. The point (-3, 2) is vertically below the center along the y-axis. The distance between (-3, 12) and (-3, 2) is 10 units. Since the radius is 9, the point (-3, 2) is outside the circle, which makes sense because the distance from the center to the point is 10, which is greater than the radius 9.

The second equation is a hyperbola. Maybe we can analyze it to see if (-3, 2) is related in some way.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So, now we have two equations:

1. The circle: \( (x + 3)^2 + (y - 12)^2 = 81 \)
2. The hyperbola: \( \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 \)

Maybe I can solve one of these equations for x or y and substitute into the other. Let's see. Let me try solving the hyperbola equation for \( (x + 3)^2 \).

<prompt_3.1> Solve the hyperbola equation for \( (x + 3)^2 \) and substitute into the circle equation. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
From the hyperbola:

\( \frac{(y - 8)^2}{9} - \frac{(x + 3)^2}{9} = 1 \)

Multiply both sides by 9:

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

Then, rearranged:

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

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

Now, substitute this into the circle equation. The circle equation is:

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

Substituting \( (x + 3)^2 \):

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

Expand the terms:

First, expand \( (y - 8)^2 \):

\( y^2 - 16y + 64 \)

Then, subtract 9:

\( y^2 - 16y + 64 - 9 = y^2 - 16y + 55 \)

Then, expand \( (y - 12)^2 \):

\( y^2 - 24y + 144 \)

So, adding these two results:

\( (y^2 - 16y + 55) + (y^2 - 24y + 144) = 81 \)

Combine like terms:

\( 2y^2 - 40y + 199 = 81 \)

Subtract 81:

\( 2y^2 - 40y + 118 = 0 \)

Divide by 2:

\( y^2 - 20y + 59 = 0 \)

Wait, this is the same equation I got earlier when subtracting the two original equations! That makes sense because subtracting the equations is equivalent to finding the radical axis, which is the line (or in this case, two horizontal lines) where the two conic sections intersect. So, solving this quadratic equation gives the y-coordinates of the intersection points.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let me solve for y:

<prompt_4.1> Solve the quadratic equation to find the y-coordinates of the intersection points. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
\( y^2 - 20y + 59 = 0 \)

Using the quadratic formula:

\( y = \frac{20 \pm \sqrt{(20)^2 - 4 \cdot 1 \cdot 59}}{2 \cdot 1} \)

Calculate discriminant:

\( 400 - 236 = 164 \)

So,

\( y = \frac{20 \pm \sqrt{164}}{2} \)

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

\( y = \frac{20 \pm 2\sqrt{41}}{2} = 10 \pm \sqrt{41} \)

So, the two y-values are \( y = 10 + \sqrt{41} \) and \( y = 10 - \sqrt{41} \).
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Now, plug these back into the expression for \( (x + 3)^2 \) from the hyperbola equation.

<prompt_5.1> Substitute the y-values back into the hyperbola equation to find the corresponding x-values. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Recall that:

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

Let's compute this for each y.

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

Compute \( (y - 8) \):

\( (10 + \sqrt{41} - 8) = 2 + \sqrt{41} \)

Square that:

\( (2 + \sqrt{41})^2 = 4 + 4\sqrt{41} + 41 = 45 + 4\sqrt{41} \)

Then subtract 9:

\( 45 + 4\sqrt{41} - 9 = 36 + 4\sqrt{41} \)

Thus, \( (x + 3)^2 = 36 + 4\sqrt{41} \)

Taking square roots:

\( x + 3 = \pm \sqrt{36 + 4\sqrt{41}} \)

Therefore, \( x = -3 \pm \sqrt{36 + 4\sqrt{41}} \)

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

Compute \( (y - 8) \):

\( (10 - \sqrt{41} - 8) = 2 - \sqrt{41} \)

Square that:

\( (2 - \sqrt{41})^2 = 4 - 4\sqrt{41} + 41 = 45 - 4\sqrt{41} \)

Subtract 9:

\( 45 - 4\sqrt{41} - 9 = 36 - 4\sqrt{41} \)

Thus, \( (x + 3)^2 = 36 - 4\sqrt{41} \)

Again, take square roots:

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

Therefore, \( x = -3 \pm \sqrt{36 - 4\sqrt{41}} \)

Therefore, the four intersection points are:

1. \( \left( -3 + \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41} \right) \)
2. \( \left( -3 - \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41} \right) \)
3. \( \left( -3 + \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41} \right) \)
4. \( \left( -3 - \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41} \right) \)
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Now, the point we need the distances to is (-3, 2). Let me denote this point as P(-3, 2). Let's compute the distance from each intersection point to P.

Take the first point: \( \left( -3 + \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41} \right) \)

The distance squared to P is:

<prompt_6.1> Calculate the distance from each intersection point to the point (-3, 2). </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
\[ \left( (-3 + \sqrt{36 + 4\sqrt{41}}) - (-3) \right)^2 + \left( 10 + \sqrt{41} - 2 \right)^2 \]

Simplify:

\[ (\sqrt{36 + 4\sqrt{41}})^2 + (8 + \sqrt{41})^2 \]

Which is:

\[ 36 + 4\sqrt{41} + 64 + 16\sqrt{41} + 41 \]

Wait, hold on. Let me compute that step by step.

First term: \( (\sqrt{36 + 4\sqrt{41}})^2 = 36 + 4\sqrt{41} \)

Second term: \( (8 + \sqrt{41})^2 = 8^2 + 2*8*\sqrt{41} + (\sqrt{41})^2 = 64 + 16\sqrt{41} + 41 = 105 + 16\sqrt{41} \)

So total distance squared is:

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

Therefore, the distance is \( \sqrt{141 + 20\sqrt{41}} \)

Similarly, for the second point: \( \left( -3 - \sqrt{36 + 4\sqrt{41}}, 10 + \sqrt{41} \right) \)

Distance squared:

\[ (-3 - \sqrt{36 + 4\sqrt{41}} + 3)^2 + (10 + \sqrt{41} - 2)^2 \]

Simplify:

\[ (-\sqrt{36 + 4\sqrt{41}})^2 + (8 + \sqrt{41})^2 \]

Which is the same as the first point:

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

So, the distance is also \( \sqrt{141 + 20\sqrt{41}} \)

Now, moving on to the third point: \( \left( -3 + \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41} \right) \)

Distance squared to P(-3, 2):

\[ (\sqrt{36 - 4\sqrt{41}})^2 + (8 - \sqrt{41})^2 \]

Compute each term:

First term: \( (\sqrt{36 - 4\sqrt{41}})^2 = 36 - 4\sqrt{41} \)

Second term: \( (8 - \sqrt{41})^2 = 64 - 16\sqrt{41} + 41 = 105 - 16\sqrt{41} \)

Adding them together:

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

So, the distance is \( \sqrt{141 - 20\sqrt{41}} \)

Similarly, the fourth point: \( \left( -3 - \sqrt{36 - 4\sqrt{41}}, 10 - \sqrt{41} \right) \)

Distance squared:

\[ (-\sqrt{36 - 4\sqrt{41}})^2 + (8 - \sqrt{41})^2 = 36 - 4\sqrt{41} + 105 - 16\sqrt{41} = 141 - 20\sqrt{41} \]

So, the distance is \( \sqrt{141 - 20\sqrt{41}} \)
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Therefore, the four distances are:

1. \( \sqrt{141 + 20\sqrt{41}} \)
2. \( \sqrt{141 + 20\sqrt{41}} \)
3. \( \sqrt{141 - 20\sqrt{41}} \)
4. \( \sqrt{141 - 20\sqrt{41}} \)
So, summing them up:

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

Hmm, that's the sum. Now, this looks like it could be simplified. Let me denote \( a = \sqrt{141 + 20\sqrt{41}} \) and \( b = \sqrt{141 - 20\sqrt{41}} \). Then, the sum is 2(a + b). Maybe we can compute (a + b)^2 to see if it simplifies.

<prompt_7.1> Simplify the expression for the sum of distances using the above plan. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Let's compute (a + b)^2:

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

Expanding this:

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

Simplify each term:

First term: 141 + 20√41

Third term: 141 - 20√41

Middle term: 2 * sqrt[ (141 + 20√41)(141 - 20√41) ]

Compute the product inside the square root:

(141)^2 - (20√41)^2 = 141² - 400*41

Calculate 141²: 141*141. Let's compute 140² = 19600, so 141² = (140 +1)^2 = 140² + 2*140*1 +1 = 19600 + 280 +1 = 19881

Then, 400*41 = 16400

Thus, the product is 19881 - 16400 = 3481

sqrt(3481) = 59, since 59² = 3481

Therefore, the middle term is 2*59 = 118

So, combining all terms:

First term + Third term = (141 + 20√41) + (141 - 20√41) = 282

Middle term: 118

Total: 282 + 118 = 400

Therefore, (a + b)^2 = 400, so a + b = 20. Since distances are positive, we take the positive square root.

Therefore, a + b = 20. Hence, 2(a + b) = 40.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
Wait, so the sum of the distances is 40? That seems neat. Let me verify the steps again to make sure I didn't make a mistake.

First, we found the intersection points by solving the two equations. Then, calculated the distances to (-3, 2), which led us to expressions involving sqrt(141 ± 20√41). Then, by recognizing that the sum of the distances can be expressed as 2(a + b), where a and b are those square roots, we computed (a + b)^2 and found it equals 400, so a + b = 20. Therefore, the total sum is 2*20 = 40.

But let me think again: the four points come in two pairs, each pair having the same distance. So each pair's distance is sqrt(141 + 20√41) and sqrt(141 - 20√41). Then, the sum is 2 times each. Then, when we added them, we recognized that the sum 2(a + b) could be simplified by squaring. But actually, the sum is 2a + 2b, which is 2(a + b). Then, we found that (a + b)^2 = 400, so a + b = 20. Therefore, 2(a + b) = 40.
Yes, that seems correct.

Therefore, the sum of the distances is 40.
That's a surprising result, but the algebra checks out. Let me just verify once more the key step where we squared the sum. If we have two terms, each appearing twice, their sum is 2*(sqrt(141 + 20√41) + sqrt(141 - 20√41)). Then, when we squared sqrt(141 + 20√41) + sqrt(141 - 20√41), we correctly expanded it and found that cross terms led to simplification, resulting in 20. Therefore, 2*20=40.
**Final Answer**
\boxed{40}
</Plan_8>

<Execution_8>
No prompt to execute.
</Execution_8>

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