SOLUTION: Find the center and radius by completing the square! x^2+y^2-4ax-6ay+4a=0

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Question 138452: Find the center and radius by completing the square!
x^2+y^2-4ax-6ay+4a=0

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Find the center and radius by completing the square!
x^2+y^2-4ax-6ay+4a=0
--------------------
x^2 -4ax + (2a)^2 + y^2 - 6ay + (3a)2 = 4a^2 + 9a^2-4a
(x-2a)^2 + (y-3a)^2 = 13a^2-4a
Center: (2a , 3a)
Radius: sqrt(13a^2-4a)
----------------------------
Cheers,
Stan H.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!
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.

Step 1: Put the constant term on the left.

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

Step 3: Divide the coefficient on the 1st degree x-term by 2 and then square the result. Add the result to both sides of the equation.

Step 4: Repeat step 3 for the y terms:

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

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

So the center of your circle is (2a,3a) and the radius is UGH!

If the original equation is really , the center comes out the same, but the radius is a nice neat