Let A be the point (1,2) and B be any point on the curve x²+y²=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 $C (α, β),$ then the length of the line segment AC is

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$\frac{4 \sqrt{5}}{5}$

$\frac{2 \sqrt{5}}{5}$

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

$\frac{6 \sqrt{5}}{5}$

answer is C.

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Detailed Solution

$A (1, 2), B (4 \cos θ, 4 \sin θ) P divides the lines segment AB in the ratio 3 : 2 \begin{array}{l} P (x, y) = \frac{2 (1, 2) + 3 (4 \cos θ, 4 \sin θ)}{3 + 2} \\ \Rightarrow 5 x - 2 = 12 \cos θ, 5 y - 4 = 12 \sin θ \\ locus of P is {(x - \frac{2}{5})}^{2} + {(y - \frac{4}{5})}^{2} = \frac{144}{25}, a circle \\ \Rightarrow C = (\frac{2}{5}, \frac{4}{5}) \\ \begin{aligned} AC = \sqrt{{(1 - \frac{2}{5})}^{2} + {(2 - \frac{4}{5})}^{2}} \\ = \sqrt{\frac{9}{25} + \frac{36}{25}} = \frac{\sqrt{45}}{5} = \frac{3 \sqrt{5}}{5} \\ ∴ AC = \frac{3 \sqrt{5}}{5} \end{aligned} \end{array}$