lf $C$ is the centre of the circle and $r$ is the radius and polar of $P$ w.r.t this circle meets $¯ ¯¯¯¯¯¯ ¯ C P$ in $Q$ then $C P . C Q =$

C P^{2} - r^{2}

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r^{2}

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C P^{2}

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P O^{2} -

r^{2}

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Solution

The correct option is B $r^{2}$
By property of pole and polor,
Let AB is a pole of a $p$ wrote on circle.
In right angled triangle ACP and ACQ,=> $r cos (90 - α) = C Q = r sin α = C Q -$ (1)
and $r = sin \times c p$ -(2)
From (1)&(2)
$C Q \times C P = r sin α \times \frac{r}{sin α}$
$C Q . C P = r^{2}$

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lf C is the centre of the circle and r is the radius and polar of P w.r.t this circle meets ¯¯¯¯¯¯¯¯CP in Q then CP. CQ=

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lf $C$ is the centre of the circle and $r$ is the radius and polar of $P$ w.r.t this circle meets $¯ ¯¯¯¯¯¯ ¯ C P$ in $Q$ then $C P . C Q =$