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. 

$$x^2+y^2−2ax+4by=0$$
Using method of completing the square 
$$x^2-2ax+(-a)^2-(-a)^2+y^2+4by+(2b)^2-(2b)^2=0$$
$$(x-a)^2+(y+2b)^2=a^2+4b^2$$
The center of the circle is given as 
$$(a,-2b)$$
The radius of the circle is 
$$r^2=a^2+4b^2$$
$$r=\sqrt{a^2+4b^2}$$
Consider the general form of equation for object circle,
$$(x-x_0)^2+(y-y_0)^2=r^2$$
Center is then
$$(x_0,y_0)$$
The radius is then
$$r$$
There is a mistake of the sign in your work.
