Let a circle of radius 4 pass through the origin O , the points $\mathrm{A}(-\sqrt{3} \mathrm{a}, 0)$ and $\mathrm{B}(0,-\sqrt{2} \mathrm{~b})$, where a and b are real parameters and $\mathrm{ab} \neq 0$. Then the locus of the centroid of $\triangle \mathrm{OAB}$ is a circle of radius

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Circle

Let a circle of radius 4 pass through the origin O , the points

A (- 3 a, 0)

and

B (0, - 2 b)

, where a and b are real parameters and

ab \neq = 0

. Then the locus of the centroid of

△ OAB

is a circle of radius

(A) $\frac{5}{3}$
(B) $\frac{7}{3}$
(C) $\frac{8}{3}$ ✓ Correct
(D) $\frac{11}{3}$

Answer: (C)

Solution

Because

O A

lies along the

x

-axis and

O B

along the

y

-axis,

∠ A O B = 9 0^{\circ}

, so

A B

is a diameter and

A B = 2 (4) = 8

. Then

(3 a)^{2} + (2 b)^{2} = 64

, i.e.

3 a^{2} + 2 b^{2} = 64

. The centroid of

△ O A B

(h, k) = (- \frac{3 a}{3}, - \frac{2 b}{3})

, so

3 a = - 3 h

and

2 b = - 3 k

, giving

3 a^{2} = 9 h^{2}

and

2 b^{2} = 9 k^{2}

. Substituting,

9 h^{2} + 9 k^{2} = 64

, so the locus is

x^{2} + y^{2} = \frac{64}{9}

, a circle of radius

\frac{8}{3}

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option

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