Q. A line is drawn through the point $P(3,11)$ to cut the circle$x^{2} + y^{2} = 9$ at $A$ and $B$. Then, $PA.PB$ is equal to
A) 9
B) 121
C) 205
D) 139
Explanation
(b)
Let $S_{1} \equiv x^{2} + y^{2} = 9\$
$PA.PB = \left( \sqrt{S_{1}} \right)^{2}$
$= (\sqrt{(3)^{2} + (11)^{2} - 9})^{2} = 121$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.