Question

What is the center and radius of the circle with equation `x^2+y^2-18x+18y=-137`?

Answer 1

The center is (9,-9) with a radius of 5

Explanation:

Rewrite the equation: `x^2+y^2-18x+18y+137=0`
The goal is to write it to something that looks like this: `(x-a)^2+(y-b)^2=r^2` where the center of the cirkel is `(a,b)` with a radius of `r`.
From looking at the coefficents of `x,x^2` we want to write: `(x-9)^2 = x^2-18x+81`
Same for `y,y^2`: `(y+9)^2=y^2+18y+81`
the part that is extra is `81 + 81 = 162 = 137 + 25`
Thus: `0=x^2+y^2-18x+18y+137=(x-9)^2+(y+9)^2 -25`
and so we find: `(x-9)^2+(y+9)^2=5^2`