<s>[INST] Find the Center and Radius of a Circle with a Given Circle Equation There is a question that is asking the center and radius of the circle that is defined by the equation $x^2 + y^2 - 2ax + 4by =0$. 
This was my work (completing the square):
$$x^2 - 2ax + (\frac{-2a}{2})^2 + y^2 + 4by + (\frac{4b}{2})^2 = (\frac{-2a}{2})^2 + (\frac{4b}{2})^2$$
$$x^2 - 2ax + a^2 + y^2 + 4by + 4b^2 = a^2 + 4b^2$$
$$(x-a)^2 + (y-2b)^2 = a^2 + 4b^2$$
I got the center of the circle to be $(a,  2b)$, and the radius of the circle to be $\sqrt{a^2 + 4b^2}$ (although I don't know how to simplify the radius any further than that). 
Am I wrong in any step? 
Thanks in advance. 
 [/INST] Note $(x - a)^2 = x^2 - 2ax + a^2$ , so $x^2 - 2ax = (x - a)^2 - a^2$. Also note that $(y + 2b)^2 = y^2 + 4by + 4b^2$, so $y^2 + 4by = (y + 2b)^2 - 4b^2$. So the equation becomes, $(x - a)^2 - a^2 + (y + 2b)^2 - 4b^2 = 0$. This implies that,
$(x - a)^2 + (y + 2b)^2 = a^2 + 4b^2$. Thus this is a circle with center $(a, -2b)$ and radius $\sqrt{a^2 + 4b^2}$.
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