A sphere of constant radius `k ,` passes through the origin and meets the axes at `A ,Ba n d Cdot` Prove that the centroid of triangle `A B C` lies on the sphere `9(x^2+y^2+z^2)=4k^2dot`

Options

A. `x^2+y^2+z^2=r^2`
B. `x^2+y^2=z^2=4r^2`
C. `9(x^2+y^2+z^2)=4r^2`
D. `3(x^2+y^2+z^2)=2r^2`

Answer

C

Explanation

Let the sphere passing through points A(a,0,0), B(0,b,0), C(0,0,c)
Equation of sphere is `x^2y^2z^2-ax-by-cz=0`
radius, `r=1/sqrt(a^2+b^2c^2)`
`rArr a^2+b^2+c^2=4r^2…(1)` (Squaring on both sides)
Let `(alpha,beta,gamma)` be centroid of sphere.
`:. (alpha,beta,gamma)=((a+0+0)/3,(0,b,0)/3,(0,0,c)/3)`
`=(a/3,b/3,c/3)`
`:. alpha^2+beta^2+gamma^2=a^2/9+b^2/9+c^2/9`
`=(a^2+b^2+c^2)/9`
`=(4r^2)/9` (form(1))
`rArr 9(alpha^2+beta^2+gamma^2)=4r^2`
So, Locus is `9(x^2+y^2+z^2)=4r^2`.