Question 24555
Hi,

The general equation of a circle as I'm sure you know is {{{(x-a)^2+(y-b)^2=r^2}}} where (a,b) is the centerpoint and r is the radius. We can make your equation look like this one by completing the square.

Firstly we're looking for the value of a such that {{{(x-a)^2=x^2+4x}}} You can either just guess that {{{a=-2}}} or expand and compare coefficients. But {{{(x-(-2))^2=x^2+4x+4}}} So {{{x^2+4x=(x+2)^2-4}}}

Doing the same for y, we get {{{y^2+2y=(y+1)^2-1}}} Finally substituting into the equation we get

{{{(x+2)^2 + (y+1)^2 -4 -1 -9=0}}}

So, in the standard form for a circle we get

{{{(x+2)^2 + (y+1)^2 =14}}}

So we can read off from this that the circle is centered at (-2,-1) with a radius of {{{sqrt(14)}}}

Hope that helps, 
Kev