$f (m_{i}, \frac{1}{m_{i}})$ , $i = 1, 2, 3, 4$ are four distinct points on the circle with centre origin, then value of $m_{1} m_{2} m_{3} m_{4}$ is equal to

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Solution

The correct option is B $1$
Equation of circle having centre of origin $(0, 0)$ and radius $= r$ ,
$S_{1}; x^{2} + y^{2} = r^{2} - - - - (1)$
$∴ f (m_{1}, \frac{1}{m_{2}}), f (m_{2}, \frac{1}{m_{2}}), - - f (m_{4}, \frac{1}{m_{4}})$
These points lie on $S_{1}$ .
Let $f (m, \frac{1}{m})$ is point lie on S_{1},
$m^{2} + \frac{1}{m^{2}} = r^{2}$
$m^{4} + 1 - r^{2} m^{2} = 0$
$m^{4} - 1 - r^{2} m^{2} + 1 = 0$
$m_{1}, m_{2}, m_{3}$ and $m_{4}$ are roots of this equation
So, $m_{1} m_{2} m_{3} m_{4} = 1$

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