Q. The greatest distance of the point $P(10,\ 7)$ from the circle $x^{2} + y^{2} - 4x - 2y - 20 = 0$ is
A) 10
B) 15
C) 5
D) None of these
Explanation
(b)
Let $S = x^{2} + y^{2} - 4x - 2y - 20$,
$S_{1} = 10^{2} + 7^{2} - 4 \times 10 - 2 \times 7 - 20 > 0$
[image:11_image29.emf]
So, $P$ lies outside the circle
Now, $PC = \sqrt{(10 - 2)^{2} + (7 - 1)^{2}} = 10$
Radius $BC = \sqrt{4 + 1 + 20} = 5$
$\therefore$ Greatest distance, $PB = PC + CB = 10 + 5$
$= 15$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.