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

1 Answer
May 9, 2018

Answer:

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#