Question

Find the center and the radius of the circle defined by the equation x^2-8x+y+12y+36=0

Answer 1

Hello,
center is (4,-6)
radius=4
since
$x^2-8x+y^2+12y+36=0\\ x^2-2*4x+16-16+y^2+2*6y+36=0\\ (x-4)^2+(y+6)^2=4^2$