Coordinate Geometry – Locus – JEE Main 24 Jan 2026 Shift 1

Solution:

Given the points $O(0,0)$, $A(-\sqrt{3}a, 0)$, and $B(0, -\sqrt{2}b)$.
Since the circle passes through O, A, and B, and $\angle AOB = 90^\circ$ (since axes are perpendicular), the segment AB must be the diameter of the circle.

Given radius $r = 4$, so diameter $AB = 8$.
Applying the distance formula for AB:
$$AB^2 = (-\sqrt{3}a – 0)^2 + (0 – (-\sqrt{2}b))^2$$
$$64 = 3a^2 + 2b^2$$

Let $G(h, k)$ be the centroid of $\Delta OAB$.
The coordinates of the centroid are given by:
$$h = \frac{-\sqrt{3}a + 0 + 0}{3} \Rightarrow 3h = -\sqrt{3}a \Rightarrow a = -\sqrt{3}h$$

$$k = \frac{0 – \sqrt{2}b + 0}{3} \Rightarrow 3k = -\sqrt{2}b \Rightarrow b = -\frac{3k}{\sqrt{2}}$$

Now, substitute the values of $a$ and $b$ into the equation $3a^2 + 2b^2 = 64$:
$$3(-\sqrt{3}h)^2 + 2\left(-\frac{3k}{\sqrt{2}}\right)^2 = 64$$

$$3(3h^2) + 2\left(\frac{9k^2}{2}\right) = 64$$

$$9h^2 + 9k^2 = 64$$

Dividing by 9:
$$h^2 + k^2 = \frac{64}{9}$$

Replacing $(h, k)$ with $(x, y)$ to find the locus:
$$x^2 + y^2 = \left(\frac{8}{3}\right)^2$$

This represents a circle with radius $R = \frac{8}{3}$.

Ans. (3)