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(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since the circle passes through the origin $O(0, 0)$ and meets the axes at $A(a, 0)$ an

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$x^2+y^2=16$

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(C) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since the circle passes through the origin $O(0, 0)$ and meets the axes at $A(a, 0)$ and $B(0, b)$, the segment $AB$ is the diameter of the circle.The length of the diameter is $2 	imes 	ext{radius} = 2 	imes 6 = 12$.Thus, $a^2 + b^2 = 12^2 = 144$.Let $(h, k)$ be the centroid of $	riangle OAB$. The coordinates of the centroid are given by $h = \frac{0+a+0}{3} = \frac{a}{3}$ and $k = \frac{0+0+b}{3} = \frac{b}{3}$.Therefore, $a = 3h$ and $b = 3k$.Substituting these into the equation $a^2 + b^2 = 144$, we get $(3h)^2 + (3k)^2 = 144$.$9h^2 + 9k^2 = 144$.Dividing by $9$, we get $h^2 + k^2 = 16$.Replacing $(h, k)$ with $(x, y)$, the locus of the centroid is $x^2 + y^2 = 16$.

If a circle of a constant radius $6$ passes through the origin $O$ and meets the coordinate axes at $A$ and $B$, then find the locus of the centroid of $\triangle OAB$.

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