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The locus of a point $P$ which divides the line joining $(1,0)$ and $(2 \cos \theta, 2 \sin \theta)$ internally in the ratio $2: 3$ for all $\theta$, is a
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The correct answer is:
Circle
Let the coordinates of the point $P$ which divides the line joining $(1,0)$ and $(2 \cos \theta, 2 \sin \theta)$ in the ratio $2: 3$
be ${ }^{(h, k)}$. Then, $h=\frac{4 \cos \theta+3}{5}$ and $k=\frac{4 \sin \theta}{5}$
$\begin{aligned}
& \Rightarrow \cos \theta=\frac{5 h-3}{4} \text { and } \sin \theta=\frac{5 k}{4} \\
& \Rightarrow\left(\frac{5 h-3}{4}\right)^2+\left(\frac{5 k}{4}\right)^2=1 \Rightarrow(5 h-3)^2+\left(5 k^2\right)=16
\end{aligned}$
Therefore locus of $(h, k)$ is $(5 x-3)^2+(5 y)^2=16$, which is a circle
be ${ }^{(h, k)}$. Then, $h=\frac{4 \cos \theta+3}{5}$ and $k=\frac{4 \sin \theta}{5}$
$\begin{aligned}
& \Rightarrow \cos \theta=\frac{5 h-3}{4} \text { and } \sin \theta=\frac{5 k}{4} \\
& \Rightarrow\left(\frac{5 h-3}{4}\right)^2+\left(\frac{5 k}{4}\right)^2=1 \Rightarrow(5 h-3)^2+\left(5 k^2\right)=16
\end{aligned}$
Therefore locus of $(h, k)$ is $(5 x-3)^2+(5 y)^2=16$, which is a circle
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