Question 138452
I did this problem on paper before I started writing the response, and I would be willing to bet that you forgot a ^2 on the 4a term because the radius comes out so horribly ugly otherwise.  I'll show you both ways, just in case.


{{{x^2+y^2-4ax-6ay+4a=0}}}


Step 1:  Put the constant term on the left.


{{{x^2+y^2-4ax-6ay=-4a}}}


Step 2:  Put the x-terms together and the y-terms together.


{{{x^2-4ax+y^2-6ay=-4a}}}


Step 3:  Divide the coefficient on the 1st degree x-term by 2 and then square the result.  {{{(4a/2)^2=(2a)^2=4a^2}}}  Add the result to both sides of the equation.


{{{x^2-4ax+4a^2+y^2-6ay=-4a+4a^2}}}


Step 4: Repeat step 3 for the y terms:  {{{(6a/2)^2=(3a)^2=9a^2}}}


{{{x^2-4ax+4a^2+y^2-6ay+9a^2=-4a+4a^2+9a^2}}}


Step 5: Collect terms on the right, and factor the two perfect squares on the left


{{{(x-2a)^2+(y-3a)^2=13a^2-4a}}}


The equation of a circle with center at (h,k) and radius r is:


{{{(x-h)^2+(y-k)^2=r^2}}}


So the center of your circle is (2a,3a) and the radius is {{{sqrt(13a^2-4a)}}} <big><b>UGH!</big></b>


If the original equation is really {{{x^2+y^2-4ax-6ay+4a^2=0}}}, the center comes out the same, but the radius is a nice neat {{{sqrt(9a^2)=3a}}}