Expert: Scott A Wilson - 2/24/2011

Question
3x^2+2x+3y^2+0y-8=0

Answer
First, the coefficient on x^2 and y^2 needs to be one, so divide the whole equation by 3.
This giveds x^2 + 2x/3 + y^2 - 8/3 = 0.

Next, note that (x+a)^2 = x^2 + 2ax + a^2, and since 2ax = 2x/3, a = 1/3.
Note that there is no y term, but only a y^2 term, so we don't have to worry about that.

To put the a^2 in, note we have to add and subtract 1/9, giving
x^2 + 2x/3 + 1/9  +  y^2  - 8 - 1/9 = 0.
Since 8 is 72/9, -8 - 1/9 is -73/9, and that can be added to both sides.
This gives (x - 1/3)^2 + y^2 = 73/9.

The standard circle is (x - x0)^2 + (y - y0)^2 = r^2,
where the center is (x0,y0) and the radius is r.

In this case, the center is at (1,3, 0) and r = root(73)/3.