Multiple choice

If a circle of constant radius $3k$ passes through the origin and meets the axes at $A$ and $B$, the locus of the centroid of $\triangle {OAB}$ is

  1. ${ x }^{ 2 }+{ y }^{ 2 }={ k }^{ 2 }\quad $
  2. ${ x }^{ 2 }+{ y }^{ 2 }=2{ k }^{ 2 }$
  3. ${ x }^{ 2 }+{ y }^{ 2 }=3{ k }^{ 2 }$

  4. None of these
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D Correct answer
AI explanation

Let the centroid of triangle OAB be (x, y, z). The vertices are O(0, 0, 0), A(a, 0, 0), and B(0, b, 0). The centroid coordinates are x = (0 + a + 0)/3 = a/3 and y = (0 + 0 + b)/0 = 0, but in 3D, the centroid of the tetrahedron is (a/4, b/4, c/4). For the triangle OAB, since the circle passes through the origin and meets the axes at A and B, the triangle is right-angled at O, and the hypotenuse AB is the diameter of the circle. The length of AB is 2 times the radius, which is 6k. Using the Pythagorean theorem, a^2 + b^2 = (6k)^2 = 36k^2. The centroid is (a/3, b/3), so a = 3x and b = 3y. Substituting these into the Pythagorean result gives (3x)^2 + (3y)^2 = 36k^2, which simplifies to x^2 + y^2 = 4k^2. Because this result does not match any of the provided options, the correct choice is None of these.

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