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If the polar of a point on the circle $x^2+y^2=p^2$ with respect to the circle $x^2+y^2=q^2$ touches the circle $x^2+y^2=r^2$, then $p, q, r$ are in
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The correct answer is:
GP
We have, polar equation with respect to $x^2+y^2=p^2$ is $x x_1+y y_1=q^2$
This equation touches the circle
$x^2+y^2-r^2=0$
$\begin{aligned} & \text { Then, } \quad r=\frac{0+0-q^2}{\sqrt{x_1^2+y_1^2}} \\ & \Rightarrow \quad-q^2=r \sqrt{x_1^2+y_1^2} \Rightarrow-q^2=r \sqrt{p^2} \\ & \Rightarrow \quad-q^2= \pm r p \Rightarrow q^2=p r \\ & \end{aligned}$
Thus $p, q$ and $r$ in GP.
This equation touches the circle
$x^2+y^2-r^2=0$
$\begin{aligned} & \text { Then, } \quad r=\frac{0+0-q^2}{\sqrt{x_1^2+y_1^2}} \\ & \Rightarrow \quad-q^2=r \sqrt{x_1^2+y_1^2} \Rightarrow-q^2=r \sqrt{p^2} \\ & \Rightarrow \quad-q^2= \pm r p \Rightarrow q^2=p r \\ & \end{aligned}$
Thus $p, q$ and $r$ in GP.
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