Given triangle $OAB$ where $O$ is the origin, $A=(0,-\sqrt{3}a)$ and $B=(-\sqrt{2}b,0)$. Let the circumradius of $\triangle OAB$ be $4$ units. If the locus of the centroid of $\triangle OAB$ is a circle, then its radius is:
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The locus of the centroid scales directly with the coordinates of the vertices, shrinking distances by a factor of $\frac{1}{3}$.
To solve this problem, we first need to understand the geometry of the triangle \( \triangle OAB \). The points given are \( O = (0, 0) \), \( A = (0, -\sqrt{3}a) \), and \( B = (-\sqrt{2}b, 0) \). We need to find the locus of the centroid of this triangle and deduce its properties.
The coordinates of the centroid \( G \) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) are given by: \(G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\).
Substituting the coordinates of \( O, A, \) and \( B \) into the formula for the centroid, we have: \(G = \left( \frac{0 - \sqrt{2}b + 0}{3}, \frac{0 - \sqrt{3}a + 0}{3} \right) = \left( -\frac{\sqrt{2}b}{3}, -\frac{\sqrt{3}a}{3} \right)\).
According to the problem, the circumradius of \( \triangle OAB \) is 4. The circumradius \( R \) for a triangle with vertices at distances \( a, b, c \) from the origin (the circumcenter coinciding with the centroid for right angle at the origin) is given by: \(R = \frac{\sqrt{OA^2 + OB^2 + AB^2}}{2}\). Here \( OA = \sqrt{3}a \), \( OB = \sqrt{2}b \), and \( AB = \sqrt{3a^2 + 2b^2} \).
Substitute these to find the expression that satisfies the condition \( R = 4 \): \(\frac{\sqrt{(\sqrt{3}a)^2 + (\sqrt{2}b)^2 + (\sqrt{3}a)^2 + (\sqrt{2}b)^2}}{2} = 4\). Simplifying, we have: \(\frac{\sqrt{6a^2 + 4b^2 + 6a^2 + 4b^2}}{2} = 4\), leading to, \(\frac{\sqrt{12a^2 + 8b^2}}{2} = 4\), thus, \(\sqrt{12a^2 + 8b^2} = 8\). Squaring both sides, we get: \(12a^2 + 8b^2 = 64\).
The locus of the centroid point \((- \frac{\sqrt{2}b}{3}, - \frac{\sqrt{3}a}{3}) \) from the relation leads us to the circle of equation: \((4)^2 = \left(\frac{\sqrt{2}b}{3}\right)^2 + \left(\frac{\sqrt{3}a}{3}\right)^2 \right)\), which simplifies to: \(\frac{2b^2}{9} + \frac{3a^2}{9} = \frac{12a^2 + 8b^2}{27} = \left(\frac{64}{27}\right)\).
After simplification, this describes a circle with radius: \(\frac{8}{3}\).
Therefore, the radius of the locus of the centroid is \(\frac{8}{3}\). The correct answer is thus:
