question_answer

If the line \[x\cos \alpha +y\sin \alpha =p\] represents the common chord of the circles \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] and \[{{x}^{2}}+{{y}^{2}}+{{b}^{2}}(a>b),\] where A and B lie on the first circle and P and Q lie on the second circle, then AP is equal to

A) \[\sqrt{{{a}^{2}}+{{p}^{2}}}+\sqrt{{{b}^{2}}+{{p}^{2}}}\]

B) \[\sqrt{{{a}^{2}}-{{p}^{2}}}+\sqrt{{{b}^{2}}-{{p}^{2}}}\]

C) \[\sqrt{{{a}^{2}}-{{p}^{2}}}-\sqrt{{{b}^{2}}-{{p}^{2}}}\]

D) \[\sqrt{{{a}^{2}}+{{p}^{2}}}-\sqrt{{{b}^{2}}+{{p}^{2}}}\]

Correct Answer: C

Solution :

[c] The given circles are concentric with centre at (0, 0) and the length of the perpendicular from (0, 0) on the given line is p. Let OL = p Then, \[AL=\sqrt{O{{A}^{2}}-O{{L}^{2}}}=\sqrt{{{a}^{2}}-{{p}^{2}}}\] and \[PL=\sqrt{O{{P}^{2}}-O{{L}^{2}}}=\sqrt{{{b}^{2}}-{{p}^{2}}}\] \[\Rightarrow AP=\sqrt{{{a}^{2}}-{{p}^{2}}}-\sqrt{{{b}^{2}}-{{p}^{2}}}\]