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

(-12,-,-,,2)

Hello :
x² + y² - 12 x - 2 y + 12 = 0
(x²-12x) +(y² -2y) +12 = 0
(x²-2(6)(x)+6²)-6² +(y² -2y+1) -1+12 = 0
(x-6)² +(y-1)² = 5²
the center of a circle is (6, 1)

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