Question 469611
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Hi,
Standard Form of an Equation of a Circle is {{{(x-h)^2 + (y-k)^2 = r^2}}}
where Pt(h,k) is the center and r is the radius
x2 + y2 - 6x - 12y + 36 = 0   |Completing the Squares
(x-3)^2 -9 + (y-6)^2 - 36 + 36 = 0
(x-3)^2 + (y-6)^2 = 9   C(3,6)  radius = 3