Circles Part 2

Concept:For a right triangle inscribed in a circle, the hypotenuse is the diameter.Here $\angle AOB = 90^\circ$, so $AB$ is the diameter of the given circle.Explanation:Circle radius $= 4$, so diameter $= 8$.Distance $AB = 8$ gives $\sqrt{(-\sqrt{3}a)^2 + (0 + \sqrt{2}b)^2} = 8$.This simplifies to $3a^2 + 2b^2 = 64$. (1)Centroid $G(h,k)$ of $\triangle OAB$: $h = \frac{0 - \sqrt{3}a + 0}{3} = -\frac{\sqrt{3}}{3}a$, $k = \frac{0+0-\sqrt{2}b}{3} = -\frac{\sqrt{2}}{3}b$.Thus $a = -\sqrt{3}h$, $b = -\frac{3k}{\sqrt{2}}$.Substitute into (1): $3(-\sqrt{3}h)^2 + 2\left(-\frac{3k}{\sqrt{2}}\right)^2 = 64$.Simplify: $9h^2 + 9k^2 = 64$, so $h^2 + k^2 = \left(\frac{8}{3}\right)^2$.Hence the locus is a circle centered at the origin with radius $\frac{8}{3}$.Answer:Radius = $\frac{8}{3}$, which corresponds to option D.