The locus of the image of the point (2, 3) in | vedclass

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(D) Let $P = (2, 3)$ be the given point and $P' = (h, k)$ be its image in the line $L_k: (2x - 3y + 4) + k(x - 2y + 3) = 0$.The line $L_k$ passes thro

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circle of radius $\sqrt{2}$

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(D) Let $P = (2, 3)$ be the given point and $P' = (h, k)$ be its image in the line $L_k: (2x - 3y + 4) + k(x - 2y + 3) = 0$.The line $L_k$ passes through the intersection of $L_1: 2x - 3y + 4 = 0$ and $L_2: x - 2y + 3 = 0$.Solving $L_1$ and $L_2$,we get $x = 1, y = 2$. Let this point be $A = (1, 2)$.Since $P'$ is the image of $P$ in $L_k$,the line $AP$ is perpendicular to $AP'$,and $AP = AP'$.The slope of $AP$ is $m_{AP} = \frac{3-2}{2-1} = 1$.Let $P' = (x, y)$. The slope of $AP'$ is $m_{AP'} = \frac{y-2}{x-1}$.Since $AP \perp AP'$,the product of their slopes is $-1$,so $\frac{y-2}{x-1} = -1 \implies y - 2 = -(x - 1) \implies x + y = 3$.Also,the distance $AP = \sqrt{(2-1)^2 + (3-2)^2} = \sqrt{2}$.Since $AP = AP'$,we have $(x-1)^2 + (y-2)^2 = AP^2 = 2$.Thus,the locus is a circle with center $(1, 2)$ and radius $\sqrt{2}$.

The locus of the image of the point $(2, 3)$ in the line $(2x - 3y + 4) + k(x - 2y + 3) = 0, k \in R$ is a:

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