Question

Show that the equation of a circle passing through the origin and having intercepts a and b on real and imaginary axes, respectively, on the argand plane is given by $z\overline{z}=a\left(Rez\right)+b\left(Imz\right).$

Answer 1

From figure,
$arg\left(\frac{z-a}{z-ib}\right)=\pm \frac{\pi }{2}$
$⟹\frac{z-a}{z-ib}+\frac{\overline{z}-a}{\overline{z}+ib}=0$
$⟹z\overline{z}-a\left(\frac{z+\overline{z}}{2}\right)-b\left(\frac{z-\overline{z}}{2i}\right)=0$
$⟹z\overline{z}=a\left(Rez\right)+b\left(Imz\right)$
Ans: 1