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 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 $(-3,2).$

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

Rather than solving the problem by eliminating x we can eliminate y. We will get a parabola which contains all the roots having x axis as directrix and (3,-2) as focus. So, all we need to do is find the sum of ordinates to find the distance. And, then you can eliminate x from equation to get quadratic in y and find sum of roots which would be doubled because of repeated ordinates.
So, ans is 40
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