Q. A line through $P(1,\ 4)$ intersect a circle $x^{2} + y^{2} = 16$ at $A$ and$B$, then$PA$. $PB$ is equal to
A) 1
B) 2
C) 3
D) 4
Explanation
(a)
Since, $PA.PB = PT^{2}$, where $PT$ is length of tangent
Here, $PT = \sqrt{S_{1}} = \sqrt{1^{2} + 4^{2} - 16} = 1$
$\therefore\ PA.PB = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.