If the equation a | z | ^ 2 + ¯ ¯ α z + α ¯ z + d = 0 represents a...

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

$∵ {\left|z\right|}^{2} = z\overset{¯}{z} \text{ and } \overset{¯}{\overset{¯}{\alpha}z + \alpha\overset{¯}{z}} = \overset{¯}{\overset{¯}{\alpha}z} + \overset{¯}{\alpha\overset{¯}{z}} = \alpha\overset{¯}{z} + \overset{¯}{\alpha}z$

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

Centre $= \dfrac{- \alpha}{a}, \text{radius,}r = \sqrt{\dfrac{\alpha\overset{¯}{\alpha}}{{a}^{2}} - \dfrac{d}{a}} = \sqrt{\dfrac{\alpha\overset{¯}{\alpha} - ad}{{a}^{2}}}$

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

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