Question

Suppose that receiving stations​ X, Y, and Z are located on a coordinate plane at the points ​(9​,5​), ​(negative 9​,negative 8​), and ​(negative 9​,2​), respectively. The epicenter of an earthquake is determined to be 10 units from X​, 13 units from​ Y, and 13 units from Z. Where on the coordinate plane is the epicenter​ located?

Answer 1

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,

$(x-9)^2 + (y-5)^2 = 100$

$(x+9)^2 + (y+8)^2 = 169$

$(x+9)^2 + (y-2)^2 = 169$

Three equations and two unknowns is an overdetermined system.  It may have no solution at all.  Let's solve the last two first.  Subtracting,

$(y+8)^2 - (y-2)^2 = 0$

and factoring as the difference of two squares gives

$(y+8-(y-2))(y+8+(y -2))=0$

$10(2y+6)=0$

$y = -3$

$(x+9)^2 + (-3+8)^2 = 169$

$(x+9)^2 = 169 - 25 = 144$

$x+9 = \pm 12$

$x = -9 \pm 12$

That's two points of intersection of the last two circles,

$(3, -3) \textrm{ and } (-21, -3)$

We need to check if either of these satisfies the first equation.  (3, -3) first.

$(3-9)^2 + (-3-5)^2 = 36 + 64= 100\quad\checkmark$

That one checks.  That's the answer.

Answer: (3,-3)