Question

A sphere of constant radius $2k$ passes through the origin and meets the axes in $A,B$ and $C$. The locus of a centroid of the tetrahedron $OABC$ is-

• A. $x^2+y^2+z^2=4k^2$
• B. $x^2+y^2+z^2=k^2$
• C. $2(k^2+y^2+z{)}^2=k^2$
• D. None of these.

Answer 1

The correct option is B $x^2+y^2+z^2=k^2$

Let the coordinate of the point $A,B,C$ be $(a,0,0,);(0,b,0);(0,0,c)$ respectively
Let this sphere also pass through the origin $(0,0,0)$.
Then the equation of the sphere $OABC$ is $x^2+y^2+z^2-ax-by-cz=0$
Radius of the sphere $={\{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2+{\left(\frac{c}{2}\right)}^2\}}^{\frac{1}{2}}\Rightarrow \frac{1}{2}\sqrt{a^2+b^2+c^2}=2k$
$\Rightarrow a^2+b^2+c^2=16k^2$
Now if $(x_1,y_1,z_1)$ is centroid of the tetrahedron with vertices as
$O(0,0,0),A(a,0,0,),B(0,b,0)$ and $C(0,0,c)$
Then we have $x_1=\frac{a+0+0+0}{4}=\frac{a}{4}$
Similarly $y_1=\frac{b}{4},z_1=\frac{c}{4}$
$\therefore a=4x_1,b=4y_1,c=4z_1$
Substituting the value of $a,b,c$
${\left(4x_1\right)}^2+{\left(4y_1\right)}^2+{\left(4z_1\right)}^2={16k}^2\Rightarrow {x_1}^2+{y_1}^2+{z_1}^2=K^2$
The locus of $(x_1,y_1,z_1)$ is ${x_1}^2+{y_1}^2+{z_1}^2=k^2$