If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle. 
If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
  Also find the center and radius of the circle.

Here's my solution, I'm not sure if it's correct or not (specifically the conditions on $A$, $B$ and $C$. I feel that my conditioning is invalid and that this may be some sort of triangle inequality case.
I grouped the $x^2$ and $Ax$ terms, and the $y^2$ and By terms together and equated them to -$C$. 
After completing the square I am left with the following expression:
$(x + A/2)^2 + (y + B/2)^2 = -C + A^2/4 - B^2/4$
Conditions on A,B and C:
****CORRECTION MADE (Redundant inequalities need not be included)****
$C < (A^2 + B^2)/4$
Radius of the circle is  $\sqrt{-C + A^2/4 + B^2/4}$
The center of the circle is $(-A/2, -B/2)$.
Correct?

Completing the square is the right thing to do. We get
$$(x + A/2)^2 + (y + B/2)^2 = -C + A^2/4 + B^2/4$$
(I have corrected a little typo in the post).
For a circle, we need to have the right-hand side positive, or if you admit degenerate circles, non-negative. The condition I would use is
$$\frac{A^2}{4}+\frac{B^2}{4}-C\gt 0,$$
or something equivalent to that, such as $A^2+B^2\gt 4C$. No other condition is needed. (Note that in the post, the inequality runs the wrong way.)
The centre is right. The radius is $\sqrt{\frac{A^2}{4}+\frac{B^2}{4}-C}$. 
