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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Solution

The correct option is C $1$
Let the eqn of circle joining these four points be, $(m_{1}, \frac{1}{m_{1}})$ where $i = 1, 2, 3, 4$ and is of form $y = \frac{1}{x}$

$x^{2} + y^{2} + 2 f y + 2 g x + c = 0$

Putting $y = \frac{1}{x}$

$x^{2} + {(\frac{1}{x})}^{2} + \frac{2 f}{x} + 2 g x + c = 6$

$x^{4} + 1 + 2 f x + 2 g x^{3} + c X^{2} = 0$

$x^{4} + 2 g x^{3} + c x^{2} + 2 f x + 1 = 0 f o r w h i c h \begin{matrix} m_{1} m_{2} m_{3} m_{4} \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ r o o t s$

So, $m_{1}, m_{2}, m_{3}, m_{4} = (\frac{+ 1}{1}) = 1$

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