Question

The radius of the circle $x^2+y^2+z^2-2y-4z-11=0$ , $x+2y+2z-15=0$ is:

• A. $\sqrt{3}$
• B. $\sqrt{5}$
• C. $\sqrt{7}$
• D. $3$

Answer 1

Answer: (C)

Radius of a circle in a sphere
$=\sqrt{\text{Radius of sphere}{)}^2-(\perp \text{distance from origin to sphere}{)}^2}$
Given equation of sphere is
$x^2+y^2+z^2-2y-4z-11=0$
Whose centre is $O(0,1,2)$ and radius $OP$ is $4$ .
$\therefore$ Distance of a plane $x+2y+2z-15=0$
from $(0,1,2)$ is
$ON=\frac{|0+2+4-15|}{\sqrt{1+4+4}}=\frac{9}{3}=3$
In $\Delta ONP$ , $O{P}^2=O{N}^2+N{P}^2$
$\Rightarrow$ $N{P}^2=(4{)}^2-(3{)}^2=7$
$\Rightarrow$ $NP=\sqrt{7}$