If $(m_{i}, 1 / m_{i}), i = 1, 2, 3, 4$ are concyclic points, then the value of $m_{1} m_{2} m_{3} m_{4} i s . . . . . . . . . . . . . . . . . .$

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Detailed Solution

Let the equation of a circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

$(m, \frac{1}{m})$ is a point on the circle $\Rightarrow$ $m^{2} + \frac{1}{m^{2}} + 2 g m + 2 f (\frac{1}{m}) + c = 0$

$\Rightarrow m^{4} + 1 + 2 g m^{3} + 2 f m + c m^{2} = 0 \Rightarrow m^{4} + 2 g m^{3} + c m^{2} + 2 f m + 1 = 0$

Let $m_{1}, m_{2}, m_{3}, m_{4}$ be the roots of this equation.

Then $(m_{i}, \frac{1}{m_{i}}), i = 1, 2, 3, 4$ one con cyclic. Product of the roots ; $m_{1} m_{2} m_{3} m_{4} = 1$ .

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