Question

Find the center of and radius of the given circle x^2+y^2-12x-2y-12=0

Answer 1

complete the squares again for the x and y terms and compare with standard equation of a circle

Answer 2

ok I got down to (x^2-6x+36)+(y^2-y+1)=12+36+1

Answer 3

x^(2)+y^(2)-12x-2y-12=0

x^(2)-12x+y^(2)-2y=12



(x-6)^(2)+(y-1)^(2)=12+36+1

(x-6)^(2)+(y-1)^(2)=49

(x-h)^(2)+(y-k)^(2)=r^(2)

Center: (6,1)_Radius: 7