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)