## Calculus 8th Edition

$a^{2}+b^{2}>4c$
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 $a^{2}+b^{2}>4c$