# Finding the distance of the intersection point of two conics to $(-3,2)$

Consider the intersection of the curves $$x^2+y^2+6x-24y+72=0, x^2-y^2+6x+16y-46=0$$. Determine the sum of the distances of their intersection points and $$(-3,2)$$.

My first thought looking at this problem was homogenization after shifting the origin to $$(-3,2)$$ but then I realized that homogenization is done with a line and a curve not with two implicit curves.

Next, I plotted these curves; the first one is a circle with center $$(-3,12)$$, radius $$r=9$$ and the second curve is a hyperbola with conjugate axis $$y=8$$ and transverse axis $$x=-3$$.

I also noticed how the point $$(-3,2)$$ lies on the transverse axis of the hyperbola and divides the line joining it, the center of the circle, and the center of hyperbola in a $$3:5$$ ratio externally. I thought of taking the parametric coordinates of the intersection points and determining relations in between individual parameters of each intersection point and using those to write the distance formula carefully but I ended up lost in a mess of 16 parameters and 8 independent relations between them.

Next, I decided to cheat a bit and used a graphing calculator to find the intersection points and then calculate the distances from $$(-3,2)$$. Interestingly, although the intersection points aren't clean simple rational numbers and the distances are even worse, the sum of all the distances is $$39.992$$ which seemed pretty close to $$40$$, the reported answer.

All in all, although this is a bit of a homework problem, I'm really curious to find an answer to this; firstly what should be my approach to these kinds of problems in which we do not need to explicitly find the points of intersection but need to calculate the distance of these points of intersection to a certain point? Secondly, is there anything particularly special about the point $$(-3,2)$$ that allows for a "clean" number to pop out?

First, standardize the equations by completing the square: $$0 = x^2 + y^2 + 6x - 24y + 72 = (x+3)^2 + (y-12)^2 - 9^2, \tag{1}$$ so the circle has center at $$(-3,12)$$ and radius $$r = 9$$; and $$0 = x^2 - y^2 + 6x + 16y - 46 = (x+3)^2 - (y-8)^2 + 3^2, \tag{2}$$ so the hyperbola has center at $$(-3, 8)$$. This, along with the "anchor" point $$(-3,2)$$ from which the distances to all intersections are measured, already suggests exploiting symmetry about the line $$x = -3$$. Eliminating $$x$$ is easy: $$(1)$$ minus $$(2)$$ implies $$0 = (y-12)^2 + (y-8)^2 - 9^2 - 3^2 = 2(y^2 - 20y + 100 - 41) \tag{3}$$ hence $$y = 10 \pm \sqrt{41}. \tag{4}$$ Next, we note that if $$(x,y)$$ is an intersection point, then the squared distance $$d^2(x,y)$$ from the anchor $$(-3,2)$$ is $$d^2(x,y) = (x+3)^2 + (y-2)^2. \tag{5}$$ Consequently, we don't need to explicitly calculate the $$x$$-value of the intersection points: instead, we rewrite either $$(1)$$ or $$(2)$$ in terms of $$d^2(x,y)$$; e.g. $$(1)$$ becomes $$0 = d^2(x,y) - (y-2)^2 + (y-12)^2 - 9^2$$ or $$d^2(x,y) = 20y - 59. \tag{6}$$ Notice how this expression doesn't depend on $$x$$ at all. For the two $$y$$-values $$(4)$$, we calculate $$d^2(x,10 \pm \sqrt{41}) = 141 \pm 20 \sqrt{41} = 100 \pm 2 \cdot 10 \sqrt{41} + (\sqrt{41})^2 = (10 \pm \sqrt{41})^2, \tag{7}$$ and recalling that the intersection points are symmetric about the line $$x = -3$$ (so there are four intersection points), the sum of distances is therefore $$2 (10 + \sqrt{41} + 10 - \sqrt{41}) = 40.$$