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

• A

  $\sqrt{a^{2}+p^{2}}+\sqrt{b^{2}+p^{2}}$
• B

  $\sqrt{a^{2}-p^{2}}+\sqrt{b^{2}-p^{2}}$
• Correct answer
  C

  $\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}}$
• D

  $\sqrt{a^{2}+p^{2}}-\sqrt{b^{2}+p^{2}}$

Solution

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 $\mathrm{PL}=\sqrt{\mathrm{OP}^{2}-\mathrm{OL}^{2}}=\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$ $\Rightarrow \mathrm{AP}=\sqrt{\mathrm{a}^{2}-\mathrm{p}^{2}}-\sqrt{\mathrm{b}^{2}-\mathrm{p}^{2}}$

Hence, the correct answer is option: 'C'.