Mathematics - Complex Number Question with Solution | TestHub

Given $a|z{|}^2+\overline{\overline{\alpha }z+\alpha \overline{z}}+d=0$

$\because {\left|z\right|}^2=z\overline{z} \& \overline{\overline{\alpha }z+\alpha \overline{z}}=\overline{\overline{\alpha }z}+\overline{\alpha \overline{z}}=\alpha \overline{z}+\overline{\alpha }z$

Then, $az\overline{z}+\alpha \overline{z}+\overline{\alpha }z+d=0$or $z\overline{z}+\frac{\alpha }{a}\overline{z}+\frac{\overline{\alpha }}{a}z+\frac{d}{a}=0$ is the equation of circle. 

Centre $=\frac{-\alpha }{a}, \text{radius, }r=\sqrt{\frac{\alpha \overline{\alpha }}{a^2}-\frac{d}{a}}=\sqrt{\frac{\alpha \overline{\alpha }-ad}{a^2}}$

${\left|\frac{\alpha }{a}\right|}^2-\frac{d}{a}\ge 0$ for the equation to represent circle.

So, $|\alpha {|}^2-ad>0$ and $a\in R-\left\{0\right\}$