Fun. Let's renotate a bit and say we have A(9,5), B(-9,-8), C(-9,2) and seek P(x,y) such that
|AP|=10, |BP|=13, |CP|=13
It's an odd fact about geometry that squared distances are often algebraically more fundamental than distance. In other words, we have
|AP|²=100, |BP|²=|CP|²=169
That tells us three equations, three circles whose meet we seek,
Three equations and two unknowns is an overdetermined system. It may have no solution at all. Let's solve the last two first. Subtracting,
and factoring as the difference of two squares gives
That's two points of intersection of the last two circles,
We need to check if either of these satisfies the first equation. (3, -3) first.
That one checks. That's the answer.
Answer: (3,-3)