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If $(m_{i}, \frac{1}{m_{i}}), i = 1, 2, 3, 4$ are concyclic points then the value of $m_{1} m_{2} m_{3} m_{4}$ is

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Detailed Solution

$x^{2} + y^{2} = a^{2} \Rightarrow$ point $(m, \frac{1}{m})$
Lies on $x^{2} + y^{2} = a^{2} \Rightarrow m^{2} + \frac{1}{m^{2}} = a^{2}$
$\Rightarrow m^{4} - a^{2} m^{2} + 1 = 0 \Rightarrow m_{1} m_{2} m_{3} m_{4} = 1$

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