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 triangle OAB is a circle of radius :

Mathematics · Circles

JEE Main 2026 — 24 January, Morning Shift — Question 2

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 :

Option A:
$\frac{5}{3}$
Option B:
$\frac{7}{3}$
Option C:
$\frac{8}{3}$
Correct
Option D:
$\frac{11}{3}$

Answer: C

Step-by-step solution

$\mathrm{AB}=8$

$3 a^{2}+2 b^{2}=64$ Centroid $\mathrm{G}(\mathrm{h}, \mathrm{k})$

$\mathrm{h}=-\frac{\sqrt{3} \mathrm{a}}{3}, \mathrm{k}=-\frac{\sqrt{2} \mathrm{~b}}{3}$

$\mathrm{a}=-\sqrt{3} \mathrm{~h}, \mathrm{~b}=\frac{-3}{\sqrt{2}} \mathrm{k}$

$9 \mathrm{~h}^{2}+9 \mathrm{k}^{2}=64$

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

$\mathrm{r}=\frac{8}{3}$

Answer key and solution verified before publishing.

Practise Circles

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Circles
Topic: Introduction to Circles