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

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Solution

The correct option is A
1

Let equation of circle be $x^{2} + y^{2}$ + 2gx + 2fy + c = 0. If $(m, \frac{1}{m})$ lies on this circle, then

$m^{2} + \frac{1}{m^{2}} + 2 g m + 2 f \frac{1}{m} + c = 0$

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

This is a fourth degree equation in m having $m_{1}, m_{2}, m_{3}, m_{4}$ as its roots.

Therefore, $m_{1} m_{2} m_{3} m_{4}$ = product of roots = $\frac{1}{1}$ = 1.

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