Let A be the point (1, 2) and B be any point | vedclass

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(D) Let $B = (4 \cos 	heta, 4 \sin 	heta)$ be any point on the circle $x^2 + y^2 = 16$.Point $P(h, k)$ divides $AB$ in the ratio $3:2$. Using the se

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$\frac{3 \sqrt{5}}{5}$

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(D) Let $B = (4 \cos 	heta, 4 \sin 	heta)$ be any point on the circle $x^2 + y^2 = 16$.Point $P(h, k)$ divides $AB$ in the ratio $3:2$. Using the section formula:$h = \frac{3(4 \cos 	heta) + 2(1)}{3 + 2} = \frac{12 \cos 	heta + 2}{5} \Rightarrow 12 \cos 	heta = 5h - 2$$k = \frac{3(4 \sin 	heta) + 2(2)}{3 + 2} = \frac{12 \sin 	heta + 4}{5} \Rightarrow 12 \sin 	heta = 5k - 4$Squaring and adding the two equations:$(12 \cos 	heta)^2 + (12 \sin 	heta)^2 = (5h - 2)^2 + (5k - 4)^2$$144 = 25(h - \frac{2}{5})^2 + 25(k - \frac{4}{5})^2$$(h - \frac{2}{5})^2 + (k - \frac{4}{5})^2 = \frac{144}{25} = (\frac{12}{5})^2$This represents a circle with centre $C(\alpha, \beta) = (\frac{2}{5}, \frac{4}{5})$.The length of $AC$ is the distance between $A(1, 2)$ and $C(\frac{2}{5}, \frac{4}{5})$:$AC = \sqrt{(1 - \frac{2}{5})^2 + (2 - \frac{4}{5})^2} = \sqrt{(\frac{3}{5})^2 + (\frac{6}{5})^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \sqrt{\frac{45}{25}} = \frac{3 \sqrt{5}}{5}$

Let $A$ be the point $(1, 2)$ and $B$ be any point on the curve $x^2 + y^2 = 16$. If the centre of the locus of the point $P$,which divides the line segment $AB$ in the ratio $3:2$ is the point $C(\alpha, \beta)$,then the length of the line segment $AC$ is

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