What is the center of a circle whose equation is x2 y2 - 12x - 2y 12 = 0?

The equation of a circle has a general form of,

(x - h) ² + (y - k) ² = r²

where the center of the circle is (h, k). Converting the above given to this form will give us,

(x - 6) ² + (y - 1) ² = 12 + 36 + 1

The value of center (h, k) is therefore (6, 1).