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If the line $x \cos \alpha+y \sin \alpha=p$ represents the common chord of the circles $x^{2}+y^{2}=a^{2}$ and $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{b}^{2}(\mathrm{a}>\mathrm{b}),$ where $\mathrm{A}$ and $\mathrm{B}$ lie ont he first circle and $\mathrm{P}$ and $\mathrm{Q}$ lie on the second circle, then AP is equal to
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The correct answer is:
$\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}}$
The given circles are concentric with centre at (0,0) and the length of the perpendicular from (0,0) on the given line is $\mathrm{p}$. Let $\mathrm{OL}=\mathrm{p}$
then, $\mathrm{AL}=\sqrt{\mathrm{OA}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}$
and $\mathrm{PL}=\sqrt{\mathrm{OP}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$
$\Rightarrow \mathrm{AP}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}-\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$
then, $\mathrm{AL}=\sqrt{\mathrm{OA}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}$
and $\mathrm{PL}=\sqrt{\mathrm{OP}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$
$\Rightarrow \mathrm{AP}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}-\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$
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