What is the center and radius of the circle with equation x2+y218x+18y=137?

1 Answer
May 9, 2018

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

Explanation:

Rewrite the equation: x2+y218x+18y+137=0
The goal is to write it to something that looks like this: (xa)2+(yb)2=r2 where the center of the cirkel is (a,b) with a radius of r.
From looking at the coefficents of x,x2 we want to write: (x9)2=x218x+81
Same for y,y2: (y+9)2=y2+18y+81
the part that is extra is 81+81=162=137+25
Thus: 0=x2+y218x+18y+137=(x9)2+(y+9)225
and so we find: (x9)2+(y+9)2=52