If $(m_{i}, 1 / m_{i}), m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle , then show that $m_{1} m_{2} m_{3} m_{4} = 1$

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Solution

Let the equation of the circle be
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Since $(m_{i}, 1 / m_{i})$ lies on it, we have
$m_{i}^{2} + \frac{1}{m_{i}^{2}} + 2 g m_{i} + \frac{2 f}{m_{i}} + c = 0$
or $m_{i}^{4} + 2 g m_{i}^{3} + c m_{i}^{2} + 2 f m_{i} + 1 = 0$
Since roots of this equation in $m_{i} a r e m_{1}, m_{2}, m_{3}, m_{4}$ we have
$m_{1} m_{2} m_{3} m_{4} = \frac{c o n s t a n t t e r m}{c o e f f . o f m_{i}^{4}} = \frac{1}{1} = 1$

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