Question 153709
Note that the equation of a circle in standard form is
{{{(x-a)^2+(y-b)^2=r^2}}}
where (a,b) is the center of the circle, r is the radius.
*******************************************************************
First write the equation in standard form
{{{x2 + y2 - 6x - 12y + 36 = 0}}}
{{{x^2-6x+y^2-12y+36=0}}}
{{{x^2-6x+3^2-3^2+y^2-12y+6^2-6^2+36=0}}}
{{{x^2-6x+3^2+y^2-12y+6^2+36-3^2-6^2=0}}}
{{{(x^2-6x+3^2)+(y^2-12y+6^2)+36-9-36=0}}}
{{{(x-3)^2+(y-6)^2-9=0}}}
{{{(x-3)^2+(y-6)^2=3^2}}}
So the radius is 3.