Question

If $z_1$ and $z_2$ lie on a circle having centre at the origin, then point of intersection of the tangents at $z_1$ and $z_2$ is

• 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 $\frac{2z_1{z}_2}{z_1+z_2}$

Let point of intersection of tangents be $z_3$


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