Question

If the line $x\cos \alpha +y\sin \alpha =p$ represents common chord $APQB$ of the concentirc circles $x^2+y^2=a^2$ and $x^2+y^2=a^2$ as shown in the figure then $AP$ is equal to

• A. $\sqrt{a^2+p^2}-\sqrt{b^2+p^2}$
• B. $\sqrt{a^2+p^2}-\sqrt{b^2-p^2}$
• C. $\sqrt{a^2-p^2}-\sqrt{b^2-p^2}$
• D. $\sqrt{a^2-p^2}-\sqrt{b^2+p^2}$

Answer 1

The correct option is C $\sqrt{a^2-p^2}-\sqrt{b^2-p^2}$

The given circles are concentric with centre at $(0,0)$
Length of perpendicular from $(0,0)$ to given line $=\frac{|\left(0\right)\cos \alpha +\left(0\right)\sin \alpha -p|}{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}=p$
$AL=\sqrt{{\left(OA\right)}^2-{\left(OL\right)}^2}=\sqrt{a^2-p^2}PL=\sqrt{{\left(OP\right)}^2-{\left(OL\right)}^2}=\sqrt{b^2-p^2}AP=AL-PL=\sqrt{a^2-p^2}-\sqrt{b^2-p^2}$
So option $C$ is correct.