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.

Question

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?

Answer 1

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 $\sqrt{\frac{A^2}{4}+\frac{B^2}{4}-C}$.

Answer 2

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$.