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  • A line segment AB of length  moves such that the points A and B remain on the periphery of a circle of<br /

A line segment AB of length \lambda moves such that the points A and B remain on the periphery of a circle of
radius \lambda. Then the locus of the point, that divides the line segment AB in the ratio 2 : 3, is a circle of radius :

Option: 1

\frac{2}{3}\lambda


Option: 2

\frac{\sqrt19}{7}\lambda


Option: 3

\frac{3}{5}\lambda


Option: 4

\frac{\sqrt19}{5}\lambda


Answers (1)

best_answer

Since OAB form equilateral \Delta

\begin{aligned} & \therefore \angle \mathrm{OAP}=60^{\circ} \\ & \mathrm{AP}=\frac{2 \lambda}{5} \\ & \cos 60^{\circ}=\frac{\mathrm{OA}^2+\mathrm{AP}^2-\mathrm{OP}^2}{2 \mathrm{OA} \cdot \mathrm{AP}} \\ & \Rightarrow \frac{1}{2}=\frac{\lambda^2+\frac{4 \lambda^2}{25}-\mathrm{OP}^2}{2 \lambda\left(\frac{2 \lambda}{5}\right)} \end{aligned}

\begin{aligned} & \Rightarrow \frac{2 \lambda^2}{5}=\lambda^2+\frac{4 \lambda^2}{25}-\mathrm{OP}^2 \\ & \Rightarrow \mathrm{OP}^2=\frac{19 \lambda^2}{25} \\ & \Rightarrow \mathrm{OP}=\frac{\sqrt{19}}{5} \lambda \end{aligned}

Therefore, locus of point P is  \begin{aligned} =\frac{\sqrt{19}}{5} \lambda \end{aligned}

Posted by

shivangi.bhatnagar

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