what is the center of a circle whose equation is x2 + y2 – 12x – 2y + 12 = 0? o (-12,-2) o (-6, -1) o (6, 1) (12, 2)

(6,1)

Step-by-step explanation:

1.) group the x and y terms (and move the constant to the other side)

x²-12x+y²-2y= -12

divide the coeficcents for the terms with just 1 or 1 x variable by 2 and then square it (and add that to both sides)

x²-12x+36+y²-2y+1= -12+36+1

factor

(x-6)²+(y-1)²= 25

the center is therefore (6,1)