Question 139392
You mean the 'equation' in standard form, I think.


The standard form of an equation of a circle with center at (h,k) and radius r is {{{(x-h)^2+(y-k)^2=r^2}}}


Notice that the terms on the left are each a squared binomial, so the trick to converting an equation in the form {{{Ax^2+By^2+Cx+Dy+C=0}}} into standard form is to complete the square on each of the variables.


{{{3x^2+3y^2-18x+6y=18}}}


Notice that all of the coefficients on this one are divisible by 3, so divide through by 3 first because we need the coefficients on the 2nd degree terms to be 1.


{{{x^2+y^2-6x+2y=6}}} is an equivalent equation and describes the same conic section.


Next rearrange the terms so that the x terms are together and the y terms are together:


{{{x^2-6x+y^2+2y=6}}}


Divide the coefficient on the 1st degree x term by 2, square the result, and add that result to both sides of the equation:  {{{(-6/2)^2=(-3)^2=9}}}


{{{x^2-6x+9+y^2+2y=6+9}}}


Do the same thing with the coefficient on the 1st degree y term:
{{{(2/2)^2=1^2=1}}}, and collect terms on the right


{{{x^2-6x+9+y^2+2y+1=6+9+1}}}
{{{x^2-6x+9+y^2+2y+1=16}}}


Now we have a sum of two perfect square trinomials:  {{{x^2-6x+9=(x-3)^2}}} and {{{y^2+2y+1=(y+1)^2}}}, so:


{{{(x-3)^2+(y+1)^2=16}}}, but {{{16=4^2}}} so:


{{{(x-3)^2+(y+1)^2=4^2}}}


Now compare this result to the standard form {{{(x-h)^2+(y-k)^2=r^2}}}.  We need to make one little adjustment to make our result look exactly like the standard form pattern:


{{{(x-3)^2+(y-(-1))^2=4^2}}}


Now we can read the coordinates of the center and the value of the radius directly.


{{{h=3}}}, {{{k=-1}}}, so the center is at (3,-1)


and the radius is {{{r=4}}}