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

May 9, 2018

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

#### Explanation:

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