Given a∣z∣2+αˉz+αzˉˉ+d=0
\because {|z|}^{2}=z\bar{z}&\bar{\bar{\alpha }z+\alpha \bar{z}}=\bar{\bar{\alpha }z}+\bar{\alpha \bar{z}}=\alpha \bar{z}+\bar{\alpha }z
Then, azzˉ+αzˉ+αˉz+d=0or zzˉ+aαzˉ+aαˉz+ad=0 is the equation of circle.
Centre =a−α,radius, r=a2ααˉ−ad=a2ααˉ−ad
∣aα∣2−ad≥0 for the equation to represent circle.
So, ∣α∣2−ad>0 and a∈R−0