Answer
$a^{2}+b^{2}>4c$
Work Step by Step
An equation of the circle with center $(h,k)$ and
radius $r$ is given as:
$(x-h)^{2}+(y-k)^{2}=r^{2}$ ...(1)
Given: $x^{2}+ y^{2}+ax +by+c= 0$ ...(2)
Need to find the conditions when the above equation will represent an equation of a circle.
Since, the radius must be greater than $0$ for an equation of a circle, i.e $r>0.$
The equation (2) can be written in the standard equation of the circle as follows:
$(x-(_\frac{a}{2})^{2}+ (y-(-\frac{b}{2}))^{2} =(\frac{\sqrt {a^{2}+b^{2}-4c}}{4})$
Compare it with equation (1), we have
center,$(h,k)=(-\frac{a}{2},-\frac{b}{2})$ and $r= (\frac{\sqrt {a^{2}+b^{2}-4c}}{4})$
Conditions must be satisfied when $r>0$
Thus,
$(\frac{\sqrt {a^{2}+b^{2}-4c}}{4})>0$
${a^{2}+b^{2}-4c}>0$
Hence, the required condition will be $a^{2}+b^{2}>4c$
