Question

$z_1$ and $z_2$ lies on the circle with centre 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. $\frac{2z_1{z}_2({\overline{z}}_2-{\overline{z}}_1)}{z_1{\overline{z}}_2-z_2{\overline{z}}_1}$
• 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 $\frac{2z_1{z}_2({\overline{z}}_2-{\overline{z}}_1)}{z_1{\overline{z}}_2-z_2{\overline{z}}_1}$



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

Substracting $\left(2\right)$ from $\left(1\right)$, we get
$2({\overline{z}}_1-{\overline{z}}_2)-z_3(\frac{{\overline{z}}_1}{z_1}-\frac{{\overline{z}}_2}{z_2})=0$
$\Rightarrow 2({\overline{z}}_2-{\overline{z}}_1)=z_3(\frac{{\overline{z}}_2}{z_2}-\frac{{\overline{z}}_1}{z_1})$
$\Rightarrow 2({\overline{z}}_2-{\overline{z}}_1)=z_3\left(\frac{z_1{\overline{z}}_2-z_2{\overline{z}}_1}{z_1{z}_2}\right)$
$\Rightarrow z_3=\frac{2z_1{z}_2({\overline{z}}_2-{\overline{z}}_1)}{z_1{\overline{z}}_2-z_2{\overline{z}}_1}$