(A) 7/3

(B) 11/3

(C) 5/3

(D) 8/3

Correct Answer: 8/3

Explanation

Let the center of the circle be $(h, k)$.

Since the circle passes through the origin:

$$ h^2 + k^2 = 16 \quad (1) $$

The circle also passes through $A(-\sqrt{3}a, 0)$:

$$ (h + \sqrt{3}a)^2 + k^2 = 16 $$

Subtracting equation (1):

$$ 2\sqrt{3}ah + 3a^2 = 0 $$

$$ h = -\frac{\sqrt{3}a}{2} $$

Similarly, since the circle passes through $B(0, -\sqrt{2}b)$:

$$ h^2 + (k + \sqrt{2}b)^2 = 16 $$

Subtracting equation (1):

$$ 2\sqrt{2}bk + 2b^2 = 0 $$

$$ k = -\frac{b}{\sqrt{2}} $$

Coordinates of centroid G of ΔOAB:

$$ G\left( \frac{0 - \sqrt{3}a + 0}{3}, \frac{0 + 0 - \sqrt{2}b}{3} \right) $$

$$ G\left( -\frac{\sqrt{3}a}{3}, -\frac{\sqrt{2}b}{3} \right) $$

From earlier results:

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

Substitute in centroid coordinates:

$$ G\left( \frac{2h}{3}, \frac{2k}{3} \right) $$

Let centroid coordinates be $(x, y)$:

$$ x = \frac{2h}{3}, \quad y = \frac{2k}{3} $$

$$ h = \frac{3x}{2}, \quad k = \frac{3y}{2} $$

Substitute in $h^2 + k^2 = 16$:

$$ \left(\frac{3x}{2}\right)^2 + \left(\frac{3y}{2}\right)^2 = 16 $$

$$ \frac{9}{4}(x^2 + y^2) = 16 $$

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

Hence, the radius of the locus is:

$$ \sqrt{\frac{64}{9}} = \boxed{\frac{8}{3}} $$

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