Multiple choice

A sphere of constant radius 'K' passes through origin and meets axes in A,B,C. The centroid of the $\triangle ABC$ lies on the sphere

  1. $9 (x^{2}+y^{2}+z^{2})=4K^{2}$
  2. $3 (x^{2}+y^{2}+z^{2})=4K^{2}$
  3. $(x^{2}+y^{2}+z^{2})=4K^{2}$
  4. $x^{2}+y^{2}-z^{2}=9K^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the sphere be x^2 + y^2 + z^2 = K^2. The intercepts are A(K,0,0), B(0,K,0), and C(0,0,K). The centroid is (K/3, K/3, K/3). Substituting this into the sphere equation: (K/3)^2 + (K/3)^2 + (K/3)^2 = 3(K^2/9) = K^2/3. This does not match the options provided; however, the centroid coordinates satisfy 9(x^2+y^2+z^2) = 9(3 * K^2/9) = 3K^2.

AI explanation

The sphere equation passing through the origin with center $(u, v, w)$ is $x^2 + y^2 + z^2 - 2ux - 2vy - 2wz = 0$. The intercepts on the axes are $A(2u, 0, 0)$, $B(0, 2v, 0)$, and $C(0, 0, 2w)$, so the centroid of triangle $ABC$ is $(2u/3, 2v/3, 2w/3)$. Substituting $x = 2u/3$, $y = 2v/3$, and $z = 2w/3$ gives $x^2 + y^2 + z^2 = 4(u^2+v^2+w^2)/9$. Because the radius $K$ satisfies $K^2 = u^2 + v^2 + w^2$, multiplying by $9$ yields the sphere $9(x^2 + y^2 + z^2) = 4K^2$.

Keep practicing — related questions