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, then the length of the line segment AC is [2023]

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Solution

Q.
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 $C (α, β)$ , then the length of the line segment AC is [2023]

1 $\frac{2 \sqrt{5}}{5}$

2 $\frac{3 \sqrt{5}}{5}$

3 $\frac{6 \sqrt{5}}{5}$

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

Ans.
(2)

$Since, P (h, k) divides A B in the ratio 3 : 2 .$

So, $\frac{12 \cos θ + 2}{5} = h and \frac{12 \sin θ + 4}{5} = k$

$\Rightarrow 12 \cos θ = 5 h - 2$ ...(i)

$and 12 \sin θ = 5 k - 4$ ...(ii)

Squaring (i) and (ii) and then adding, we get

$144 = {(5 h - 2)}^{2} + {(5 k - 4)}^{2}$

$\Rightarrow {(x - \frac{2}{5})}^{2} + {(y - \frac{4}{5})}^{2} = \frac{144}{25}$

Centre $\equiv (\frac{2}{5}, \frac{4}{5}) \equiv C (α, β)$

$A C = \sqrt{{(1 - \frac{2}{5})}^{2} + {(2 - \frac{4}{5})}^{2}} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \frac{3 \sqrt{5}}{5}$

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