Answer
As long as $(\frac{a^2}{4}+\frac{b^2}{4} -c) \gt 0$, then this is the equation of a circle.
The center of the circle is $(-\frac{a}{2}, -\frac{b}{2})$
The radius is $\sqrt{\frac{a^2}{4}+\frac{b^2}{4} -c}$
Work Step by Step
We can write the general equation for a circle:
$(x-u)^2+(y-v)^2 = r^2$
where $(u,v)$ is the center of the circle and $r$ is the radius
We can rearrange the equation given in the question:
$x^2+y^2+ax+by+c = 0$
$(x^2+ax)+(y^2+by)+c = 0$
$(x+\frac{a}{2})^2-\frac{a^2}{4}+(y+\frac{b}{2})^2-\frac{b^2}{4}+c = 0$
$(x+\frac{a}{2})^2+(y+\frac{b}{2})^2 = \frac{a^2}{4}+\frac{b^2}{4} -c$
As long as $(\frac{a^2}{4}+\frac{b^2}{4} -c) \gt 0$, then this is the equation of a circle.
The center of the circle is $(-\frac{a}{2}, -\frac{b}{2})$
The radius is $\sqrt{\frac{a^2}{4}+\frac{b^2}{4} -c}$
