The centroid of a variable triangle ABC is at | vedclass

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(C) Let the coordinates of $C$ be $(h, k)$.The centroid $G$ of triangle $ABC$ is given by $\left( \frac{2+3+h}{3}, \frac{3+2+k}{3} \right) = \left( \f

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a circle of diameter $30$ units

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(C) Let the coordinates of $C$ be $(h, k)$.The centroid $G$ of triangle $ABC$ is given by $\left( \frac{2+3+h}{3}, \frac{3+2+k}{3} ight) = \left( \frac{5+h}{3}, \frac{5+k}{3} ight)$.Given that the distance of the centroid from the origin is $5$ units,we have $\sqrt{\left( \frac{5+h}{3} ight)^2 + \left( \frac{5+k}{3} ight)^2} = 5$.Squaring both sides,we get $\left( \frac{5+h}{3} ight)^2 + \left( \frac{5+k}{3} ight)^2 = 25$.Multiplying by $9$,we get $(h+5)^2 + (k+5)^2 = 225$.Replacing $(h, k)$ with $(x, y)$,the locus of $C$ is $(x+5)^2 + (y+5)^2 = 15^2$.This represents a circle with radius $r = 15$ units.The diameter of the circle is $2r = 2 	imes 15 = 30$ units.

The centroid of a variable triangle $ABC$ is at a distance of $5$ units from the origin. If $A = (2, 3)$ and $B = (3, 2)$,then the locus of $C$ is

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