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If the line \mathrm{x \cos \alpha+y \sin \alpha=p}  is a common chord of circles \mathrm{x^2+y^2=a^2, x^2+y^2=b^2}, \mathrm{(a>b)} intersecting the

outercircle at A, B and the inner circle at P, Q, then length AP =

Option: 1

\mathrm{\sqrt{a^2+p^2}+\sqrt{b^2+p^2}}


Option: 2

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


Option: 3

\mathrm{\sqrt{a^2-p^2}-\sqrt{b^2-p^2}}


Option: 4

\mathrm{2 \sqrt{a^2-b^2}}


Answers (1)

best_answer

The circles are concentric with centre at (0, 0). As the equation of the chord AB is \mathrm{x \cos \alpha+y \sin \alpha=p}, 

\mathrm{\rightarrow } the perpendicular distance from O on AB = p

\mathrm{\text { In } \triangle O M P, P M^2=b^2-p^2 \Rightarrow P M=\sqrt{b^2-p^2}}

\mathrm{\begin{aligned} & \text { In } \triangle O M A, A M^2=a^2-p^2 \Rightarrow A M=\sqrt{a^2-p^2} \\ & \therefore A P=A M-P M \end{aligned}}

\mathrm{=\sqrt{a^2-p^2}-\sqrt{b^2-p^2}}

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