Let a circle of radius 4 pass through the ori | vedclass

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(C) The circle passes through the origin $O(0,0)$, $A(-\sqrt{3}a, 0)$, and $B(0, -\sqrt{2}b)$.Since the circle passes through the origin, its equation

Vedclass · Accepted Answer

$\frac{8}{3}$

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(C) The circle passes through the origin $O(0,0)$, $A(-\sqrt{3}a, 0)$, and $B(0, -\sqrt{2}b)$.Since the circle passes through the origin, its equation is of the form $x^2 + y^2 + 2gx + 2fy = 0$.Substituting $A(-\sqrt{3}a, 0)$ into the equation: $(-\sqrt{3}a)^2 + 2g(-\sqrt{3}a) = 0 \implies 3a^2 - 2g\sqrt{3}a = 0 \implies 2g = \sqrt{3}a$.Substituting $B(0, -\sqrt{2}b)$ into the equation: $(-\sqrt{2}b)^2 + 2f(-\sqrt{2}b) = 0 \implies 2b^2 - 2f\sqrt{2}b = 0 \implies 2f = \sqrt{2}b$.The equation of the circle is $x^2 + y^2 + (\sqrt{3}a)x + (\sqrt{2}b)y = 0$.The radius of the circle is $R = \sqrt{g^2 + f^2} = 4$.Thus, $g^2 + f^2 = 16 \implies (\frac{\sqrt{3}a}{2})^2 + (\frac{\sqrt{2}b}{2})^2 = 16 \implies \frac{3a^2}{4} + \frac{2b^2}{4} = 16 \implies 3a^2 + 2b^2 = 64$.Let the centroid of $\Delta OAB$ be $G(h, k)$.$h = \frac{0 - \sqrt{3}a + 0}{3} = -\frac{\sqrt{3}a}{3} \implies a = -\sqrt{3}h$.$k = \frac{0 + 0 - \sqrt{2}b}{3} = -\frac{\sqrt{2}b}{3} \implies b = -\frac{3k}{\sqrt{2}}$.Substituting $a$ and $b$ into $3a^2 + 2b^2 = 64$:$3(-\sqrt{3}h)^2 + 2(-\frac{3k}{\sqrt{2}})^2 = 64 \implies 3(3h^2) + 2(\frac{9k^2}{2}) = 64 \implies 9h^2 + 9k^2 = 64 \implies h^2 + k^2 = \frac{64}{9}$.The locus is $x^2 + y^2 = (\frac{8}{3})^2$, which is a circle of radius $\frac{8}{3}$.

Let a circle of radius $4$ pass through the origin $O$, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab \neq 0$. Then the locus of the centroid of $\Delta OAB$ is a circle of radius

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