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`

Explanation

Let the equation of any sphere passing through the origin and having radius k be
`" "x^(2)+y^(2)+z^(2)+2ux+2vy+2wz=0" "` (i)
As the radius of the sphere is k, we get
`" "u^(2)+v^(2)+w^(2)=k^(2)" "`(ii)
Note that (i) meets the x-axis at `O(0, 0, 0) and A(-2u, 0, 0), ` y-axis at `O(0, 0, 0) and B(0, -2v, 0),` and z-axis at `O(0, 0, 0) and C(0, 0, -2w)`.
Let the centroid of the triangle `ABC` be `(alpha, beta, gamma)`. Then
`" "alpha =-(2u)/(3), beta=-(2v)/(3), gamma =- (2w)/(3) rArr" " u =- (3alpha)/(2), v =- (3beta)/(2), w =- (3v)/(2)`
Putting this in (ii), we get
`" "((-3)/(2)alpha)^(2)+((-3)/(2)beta)^(2) + ((-3)/(2)gamma )^(2) = k^(2)`
or `" "alpha^(2)+beta^(2)+gamma^(2)=(4)/(9) k^(2)`
This shows that the centroid of triangle `ABC` lies on `x^(2)+y^(2)+z^(2)=(4)/(9) k^(2)`.