$P$ and $Q$ are points on the line $y = m x + c$ . Then the polars of $P$ and $Q$ w.r.t the circle $x^{2} + y^{2} = a^{2}$ meet at the point.

(\frac{a^{2} m}{c}, \frac{a^{2}}{c})

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(\frac{a^{2} m}{c}, \frac{- a^{2}}{c})

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(\frac{- a^{2} m}{c}, \frac{a^{2}}{c})

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(\frac{- a^{2} m}{c}, \frac{- a^{2}}{c})

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Solution

The correct option is B $(\frac{- a^{2} m}{c}, \frac{a^{2}}{c})$
Let, P $(h, k)$ and Q $(r, s)$ are on line $y = m x + c$
So, $k = m h + c$ ---(1)
$s = r m + c$ ----(2)
So, Pole of P is
$x h + y k = 0^{2}$
then Q lie on $x h + y k = 0^{2}$
So, $r h + s k = a^{2}$ ---(3)
and pole of Q is
$r x + s y = a^{2}$
then P i'e on $r x + s y = a^{2}$
So, $r h + s k = a^{2}$ ---(4)
So, intersection of polar of P and Q is
$y r k - s h y = a^{2} (r - 5)$
$\Rightarrow y = \frac{a^{2} (r - 5)}{r k - h}$ and $x = \frac{a^{2} - \frac{a^{2} k (r - 5)}{r k - h s}}{h}$
$y = \frac{- a^{2}}{c}$
$= \frac{- h g a^{2} + s k a^{2}}{4}$
$x = a^{2} \frac{(s k - h 5)}{4}$
$= a^{2} s \frac{(k - 4)}{4}$
$= \frac{- a^{2} m}{c}$
So, $(\frac{- a^{2} m}{c}, \frac{a^{2}}{c})$

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