Conic-Sections Question 543
Question: 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
Options:
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}} $
Show Answer
Answer:
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{OA^{2}-OL^{2}}=\sqrt{a^{2}-p^{2}} $ and $ PL=\sqrt{OP^{2}-OL^{2}}=\sqrt{b^{2}-p^{2}} $
$ \Rightarrow AP=\sqrt{a^{2}-p^{2}}-\sqrt{b^{2}-p^{2}} $
BETA
