Question

The sum of the longest and shortest distances from the point $(1,2,-1)$ to the surface of the sphere $x^2+y^2+z^2=24$ is

• A. $3\sqrt{6}$
• B. $4\sqrt{6}$
• C. $\sqrt{6}$
• D. $2$

Answer 1

The correct option is B $4\sqrt{6}$

Centre of the sphere is $C(0,0,0)$ and radius is $r=\sqrt{24}=2\sqrt{6}$
Let given point be $P(1,2,-1)$

$\therefore CP=\sqrt{(1-0{)}^2+(2-0{)}^2+(-1-0{)}^2}=\sqrt{6}$
Hence, longest distance of P from the sphere is $r+CP=3\sqrt{6}$
and shortest distance of P from the sphere is $r-CP=\sqrt{6}$

Therefore, sum of the longest and shortest distances is $4\sqrt{6}$.