P and Q are the points of trisection of the l | vedclass

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(A) The points of trisection $P$ and $Q$ of the line segment joining $(3, -7)$ and $(-5, 3)$ are calculated using the section formula: $P = \left( \fr

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a circle with radius $\frac{\sqrt{41}}{3}$

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(A) The points of trisection $P$ and $Q$ of the line segment joining $(3, -7)$ and $(-5, 3)$ are calculated using the section formula: $P = \left( \frac{1 	imes (-5) + 2 	imes 3}{1+2}, \frac{1 	imes 3 + 2 	imes (-7)}{1+2} ight) = \left( \frac{1}{3}, -\frac{11}{3} ight)$ $Q = \left( \frac{2 	imes (-5) + 1 	imes 3}{2+1}, \frac{2 	imes 3 + 1 	imes (-7)}{2+1} ight) = \left( -\frac{7}{3}, -\frac{1}{3} ight)$ Since $PQ$ subtends a right angle at $R$,the locus of $R$ is a circle with $PQ$ as the diameter. The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting the coordinates of $P$ and $Q$: $(x - \frac{1}{3})(x + \frac{7}{3}) + (y + \frac{11}{3})(y + \frac{1}{3}) = 0$ $x^2 + 2x - \frac{7}{9} + y^2 + 4y + \frac{11}{9} = 0$ $x^2 + y^2 + 2x + 4y + \frac{4}{9} = 0$ The radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$. Here,$g = 1$,$f = 2$,and $c = \frac{4}{9}$. Radius $= \sqrt{1^2 + 2^2 - \frac{4}{9}} = \sqrt{5 - \frac{4}{9}} = \sqrt{\frac{41}{9}} = \frac{\sqrt{41}}{3}$.

$P$ and $Q$ are the points of trisection of the line segment joining the points $(3, -7)$ and $(-5, 3)$. If $PQ$ subtends a right angle at a variable point $R$,then the locus of $R$ is

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