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?
 A: 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 $\sqrt{\frac{A^2}{4}+\frac{B^2}{4}-C}$. 
A: Lets' solve for:
X^2 + y^2 + ax + by + c = 0
Step 1: Add -x^2 to both sides
ax + by + x^2 + y^2 + c + - x^2 = 0 + - x^2
ax + by + y^2 + c = - x^2
Step 2: Add - y^2 to both sides
ax + by + y^2 + c +  - y^2 = - x^2 + y^2
ax + by + c = - x^2 - y^2
Step 3: Add - by to both sides
ax + by + c + by = - x^2 - y^2 +  - by
ax + c = - by - x^2 - y^2
Step 4: Add - c to both sides
ax + c + - c = by - x^2 - y^2 +  - c
ax = - by - x^2 - y^2 - c
Step 5: Divide both sides by x.
ax/x = - by - x^2 - y^2 - c/ x
a = - by - x^2 - y^2 - c/x
Answer is:
a = - by - x^2 - y^2 - c/x
A: Your answer is almost complete.
The only thing that you are missing is actually answer the question.
For the equation to represent a circle, is needed that the radius is positive. Keep it going from $ \sqrt{ -C +\frac{A^2}{4}+\frac{B^2}{4}}$, by simplifying it: $$\sqrt{\frac{A^2+B^2-4AC}{4}} >0 \Rightarrow $$
$$\Rightarrow A^2 + B^2 - 4AC > 0$$
So, the final answer is: the condition on A,B and C such that this represents the equation of a circle is that $ A^2 + B^2 - 4AC > 0$.
