If $z_{1}, z_{2}, z_{3}$ be the complex numbers representing the points A, B, C such that $\frac{2}{z_{1}} = \frac{1}{z_{2}} + \frac{1}{z_{3}}$ , prove that a circle through A, B, C passes through the origin.

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$\frac{1}{z_{1}} + \frac{1}{z_{3}} = \frac{2}{z_{2}}$

$\frac{1}{z_{1}} + \frac{1}{z_{3}} = \frac{1}{z_{2}}$

$\frac{1}{z_{1}} - \frac{1}{z_{3}} = \frac{2}{z_{2}}$

$\frac{1}{z_{1}} + \frac{1}{z_{3}} = \frac{4}{z_{2}}$

answer is D.

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Detailed Solution

Let angles at $A and B be α and β .$

Anti-clockwise rotation about A and C gives

$- z_{1} = (z_{2} - z_{1}) e^{i α} and (z_{2} - z_{3}) = - z_{3} e^{i β}$

Multiplying, we get

$z_{1} z_{2} - z_{1} z_{3} = (z_{2} z_{3} - z_{1} z_{3}) e^{i (α + β)}$

If the points are concyclic, then $α + β = π$

Thus, $z_{1} z_{2} - z_{1} z_{3} = (z_{2} z_{3} - z_{1} z_{3}) (- 1)$

$\begin{array}{l} \Rightarrow z_{1} z_{3} - z_{1} z_{3} = (- z_{2} z_{3} + z_{1} z_{3}) \\ \Rightarrow z_{1} z_{2} + z_{2} z_{3} = 2 z_{1} z_{3} \\ \Rightarrow \frac{z_{1} z_{2}}{z_{1} z_{2} z_{3}} + \frac{z_{2} z_{3}}{z_{1} z_{2} z_{3}} = \frac{2 z_{1} z_{3}}{z_{1} z_{2} z_{3}} \\ \Rightarrow \frac{1}{z_{3}} + \frac{1}{z_{1}} = \frac{2}{z_{2}} \\ \Rightarrow \frac{1}{z_{1}} + \frac{1}{z_{3}} = \frac{2}{z_{2}} \end{array}$