Question 388025
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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
x^2+y^2-10x+12y+12=0
completing the squares to put into the Standard Form
x^2 -10x + y^2 +12y + 12 = 0
 (x-5)^2 - 25 + (y+6)^2 - 36 + 12 = 0
  (x-5)^2 + (y+6)^2 -49 = 0
  (x-5)^2 + (y+6)^2 = 49 
Center (5,-6)  radius = 7
{{{drawing(500,500, -15,15,-15,15,grid(1),circle(5,-6,7), circle(5,-6,.15))}}}