Here's the question you clicked on:
LaddiusMaximus
Find the center of and radius of the given circle x^2+y^2-12x-2y-12=0
complete the squares again for the x and y terms and compare with standard equation of a circle
ok I got down to (x^2-6x+36)+(y^2-y+1)=12+36+1
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