Let tangents at $A (z_{1}) and B (z_{2})$ are drawn to the circle $| z | = 2$ . Then which of the following is INCORRECT?

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Slope of tangent at $A (z_{1}) i s \frac{1}{i} (\frac{z_{1} + {\bar{z}}_{1}}{z_{1} - {\bar{z}}_{1}})$

The equation of tangent at A is given by $\frac{z}{z_{1}} + \frac{\bar{z}}{z_{1}} = 2$ .

If tangents at $A (z_{1}) a n d B (z_{2})$ intersect at $P (z_{p})$ , then $z_{n} = \frac{2 z_{1} z_{2}}{z_{1} + z_{2}}$ .

If points $A (z_{1}) a n d B (z_{2})$ on the circle $| z | = 2$ are such that $z_{1} + z_{2} = 0$ , then tangents intersect at $\frac{π}{2}$ .

answer is D.

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Detailed Solution

A) $\arg (\frac{z - z_{1}}{0 - z_{1}}) = \frac{π}{2}$
Question Image
$\Rightarrow \frac{z_{1} - z}{z_{1}}$ is purely imaginary.
So, $\frac{z_{1} - z}{z_{1}} + \frac{{\bar{z}}_{1} - \bar{z}}{{\bar{z}}_{1}} = 0$
$\Rightarrow \frac{z}{z_{1}} + \frac{\bar{z}}{{\bar{z}}_{1}} = 2$
B) By applying rotation, we get
$\frac{z - z_{1}}{0 - z_{1}} = \frac{p}{r} e^{\frac{i π}{2}} ... (i)$
Also $\frac{0 - z_{2}}{z - z_{2}} = \frac{r}{p} e^{\frac{i π}{2}} ... (ii)$
$∴$ On multiplying eqns. (i) and (ii), we get
$z_{p} = \frac{2 z_{1} z_{2}}{z_{1} - z_{2}}$
C) As equation of tangent at $A (x_{1}, y_{1})$ is $x x_{1} + y y_{1} = 4$
$∴$ Slope of tangent $\frac{- x_{1}}{y_{1}} = \frac{- 2 x_{1}}{2 y_{1}} = - i (\frac{z_{1} + {\bar{z}}_{1}}{z_{1} - {\bar{z}}_{1}}) = \frac{1}{i} (\frac{z_{1} + {\bar{z}}_{1}}{z_{1} - {\bar{z}}_{1}})$
D) Clearly, tangents are parallel lines.
(As $A (z_{1})$ and $B (z_{2})$ are ends of diameter of circle.)