The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points

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 of these four points from the point $(-3,2).$

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$x^2+y^2+6x-24y+72=0$ is a circle and $x^2-y^2+6x+16y-46=0$ is a hyperbola.These cut at four points.
$x^2+y^2+6x-24y+72=0$ becomes $(x+3)^2+(y-12)=9^2$.
$x^2-y^2+6x+16y-46=0$ becomes $(x+3)^2-(y-8)^2=-9$.

When i solved these two equations to find the points of intesection,$y=10\pm \sqrt{41}$
and $x$ is $-3\pm\sqrt{36\pm 4\sqrt{41}}$

Now it is very difficult to find the sum of distances of these points from $(-3,2)$.Answer is $40$ given in the book.

Answer 1

We don't have to find the coordinates of the four points.

Since $$(x^2+y^2+6x-24y+72)-(x^2-y^2+6x+16y-46)=0$$ $$\Rightarrow 2y^2-40y+118=0$$

we can set the four points as $(-3\pm\alpha,\beta),(-3\pm\gamma,\omega)$ where $$\beta+\omega=-(-40)/2=20,\qquad\beta\omega=118/2=59.$$

Now, noting that $$(-3+\alpha)^2+\beta^2+6(-3+\alpha)-24\beta+72=0\Rightarrow \alpha^2+(\beta-2)^2=20\beta-59$$

the sum of the distances can be represented as $$2\sqrt{\alpha^2+(\beta-2)^2}+2\sqrt{\gamma^2+(\omega-2)^2}=2\left(\sqrt{20\beta-59}+\sqrt{20\omega-59}\right)\tag1$$

Then, $$\begin{align}&\left(\sqrt{20\beta-59}+\sqrt{20\omega-59}\right)^2\\&=20(\beta+\omega)-2\cdot 59+2\sqrt{20^2\beta\omega-20\cdot 59(\beta+\omega)+59^2}\\&=20\cdot 20-2\cdot 59+2\sqrt{20^2\cdot 59-20\cdot 59\cdot 20+59^2}\\&=20^2\end{align}$$

The result follows from $(1)$.

Answer 2

note: for any point of 4, the distance to $(-3,2)$ is $\sqrt{(-3\pm\sqrt{36\pm 4\sqrt{41}}-(-3))^2+(10\pm \sqrt{41}-2)^2}=\sqrt{36\pm 4\sqrt{41}+(8\pm \sqrt{41})^2}=\sqrt{(10\pm\sqrt{41})^2}=10\pm\sqrt{41}$

so it is easy to see the final answer.