Question 19912
You need to get your equation into the standard form for a circle with center at (h, k) and radius r. {{{(x-h)^2 + (y-k)^2 = r^2}}}You can do this by completing the square in the x-terms and in the y-terms.

{{{x^2+y^2+Ax+By+C=0}}} Group the x-terms and the y-terms.
{{{(x^2+Ax) + (y^2+By)+C=0}}} Now subtract C from both sides of the equation.
{{{(x^2+Ax) + (y^2+By) = -C}}} Complete the square in the x-terms and in the y-terms.
{{{(x^2+Ax+A^2/4) + (y^2+By+B^2/4) = A^2/4 + B^2/4 - C}}} Factor the x-group and the y-group and simplify.
{{{(x+A/2)^2 + (y+B/2)^2 = (A^2+B^2-4C)/4}}} Compare this with:
{{{(x-h)^2 + (y-k)^2 = r^2}}}

{{{h = -A/2}}}
{{{k = -B/2}}}
{{{r = (1/2)sqrt(A^2+B^2-4C)}}}