Question

$z_1$ and $z_2$ lie on a circle with center at the origin. The point of intersection $z_3$ of the tangents at $z_1$ and $z_2$ is given by

• A. $\frac{1}{2}(\overline{z_1}+\overline{z_2})$
• B. $\frac{2z_1{z}_2}{z_1+z_2}$
• C. $\frac{1}{2}(\frac{1}{z_1}+\frac{1}{z_2})$
• D. $\frac{z_1+z_2}{\overline{z_1}\overline{z_2}}$

Answer 1

The correct option is B $\frac{2z_1{z}_2}{z_1+z_2}$

As $\Delta OAC$ is a right-angled triangle with right angle at A, so
$|z_1{|}^2+|z_3-z_1{|}^2=|z_3{|}^2$
$\Rightarrow 2|z_1{|}^2={\overline{z}}_3{z}_1-{\overline{z}}_1{z}_3=0$
$\Rightarrow 2{\overline{z}}_1-{\overline{z}}_3-\frac{{\overline{z}}_1}{z_1}{z}_3=0$ (1)
Similarly,
$2{\overline{z}}_2-{\overline{z}}_3-\frac{{\overline{z}}_2}{z_2}{z}_3=0$ (2)
$2({\overline{z}}_2-{\overline{z}}_1)=z_3(\frac{{\overline{z}}_1}{z_1}-\frac{{\overline{z}}_2}{z_2})$
$\Rightarrow \frac{2r^2(z_1-z_2}{z_1{z}_2}=z_3{r}^2\left(\frac{z_1^2-z_1^2}{z_1^2{z}_2^2}\right)[\because |z_1{|}^2=|z_2{|}^2=r^2]$
$\Rightarrow z_3=\frac{2z_1{z}_2}{z_2+z_1}$