Question

Tangent drawn to circle $\mid z\mid =2$ at $A\left(z_1\right)$ and normal at $B\left(z_2\right)$ meet at the point $P\left(z_p\right)$, then

• A.
  $zp=\frac{2\mid z_{1}mi{d}^2{z}_2}{z_1{\overline{z}}_2+\overline{z_1}{z}_2}$
• B. $zp=\frac{2\mid z_1{\mid }^2{z}_1}{z_1{\overline{z}}_2+\overline{z_1}{z}_2}$
• C.
  $zp=\frac{2\mid z_1{\mid }^2{z}_2}{z_1{\overline{z}}_2+\overline{z_1}{z}_2}$
• D. none of these

Answer 1

The correct option is C

$zp=\frac{2\mid z_1{\mid }^2{z}_2}{z_1{\overline{z}}_2+\overline{z_1}{z}_2}$
Here arg $\left(\frac{0-z_1}{z_p-z_1}\right)=\pm \frac{\pi }{2}$
$\Rightarrow \frac{z_1}{z_1-z_p}+\frac{{\overline{z}}_1}{z_1-z_p}=0..\left(i\right)$.

Also,points B,O and P are collinear, hence arg$\frac{z_2}{z_p}=\pi or\frac{z_2}{z_p}=\frac{{\overline{z}}_2}{{\overline{z}}_p}...\left(iii\right)$

from (i) and (ii),we get $Z_p=\frac{2\mid z_1{\mid }^2{z}_2}{z_1{\overline{z}}_2+\overline{z_1}{z}_2}.$