If tangents are drawn to the circle $| z | = 2$ at $A (z_{1}) a n d B (z_{2})$ ,then which of the following option(s) is/are correct?

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

If points $A (z_{1})$ and $B (z_{2})$ on the circle $| z | = 2$ are such that $z_{1} + z_{2} = 0$ , then the angle between tangents at A and B is $\frac{π}{2}$

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

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

answer is A, B.

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Detailed Solution

$A) \arg (\frac{z - z_{1}}{0 - z_{1}}) = \frac{π}{2}$

$\Rightarrow \frac{z_{1} - z}{z_{1}}$ 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
$\frac{z - z_{1}}{0 - z_{1}} = \frac{p}{r} e^{\frac{i π}{2}}$
Also $\frac{0 - z_{2}}{z - z_{2}} = \frac{r}{p} e^{\frac{i π}{2}}$