SOLUTION: Find the center of the circle. x^2+y^2-4x-12y+30=0

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find the center of the circle. x^2+y^2-4x-12y+30=0      Log On


   

Question 40200: Find the center of the circle.
x^2+y^2-4x-12y+30=0
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
We complete the square twice...once for x and once for y...we get
x^2 + y^2 - 4x - 12y + 30 = 0
x^2 - 4x + y^2 - 12y + 30 = 0
x^2 - 4x + 4 + y^2 - 12y + 36 = -30 + 4 + 36
(x - 2)^2 + (y - 6)^2 = 10
center at (2, 6)
radius is sqrt(10)