Question

A sphere of constant radius $k$ passes through the origin and meets the axes at $A,B$ and $C,$ then the centroid of $\Delta ABC$ lies on the sphere

• A. $x^2+y^2+z^2=\frac{k^2}{9}$
• B. $x^2+y^2+z^2=\frac{2k^2}{9}$
• C. $x^2+y^2+z^2=\frac{k^2}{3}$
• D. $x^2+y^2+z^2=\frac{4k^2}{9}$

Answer 1

The correct option is D $x^2+y^2+z^2=\frac{4k^2}{9}$

Equation of sphere passing through origin is :
$x^2+y^2+z^2+2ux+2vy+2wz=0$
Now, $k^2=u^2+v^2+w^2\cdots \left(i\right)$
and the sphere meets the $x-$axis at : put $y=z=0\Rightarrow A(-2u,0,0)$
Similarly $B(0,-2v,0),C(0,0,-2w)$
then centroid of $\Delta ABC$ is : $(x,y,z)=(-\frac{2u}{3},-\frac{2v}{3},-\frac{2w}{3})$
putting in $\left(i\right),$ we get
${(-\frac{3x}{2})}^2+{(-\frac{3y}{2})}^2+{(-\frac{3z}{2})}^2=k^2\Rightarrow x^2+y^2+z^2=\frac{4k^2}{9}$