Question 606829
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Hi,
x^2 + y^2 + 14x + 12y + 84=0 
x^2 + 14x + 49 - 49 + y^2 + 12y +36 -36 + 84=0
 (x+7)^2 - 49 + (y+6)^2 -36 + 84=0 
 (x+7)^2 + (y+6)^2= 1   Center(-7,-6) and radius of 1
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