The line joining (5,0) to $((10 \cos \theta, 10 \sin \theta)$ is divided internally in the ratio 2: 3 at $P$…
The line joining (5,0) to $((10 \cos \theta, 10 \sin \theta)$ is divided internally in the ratio 2: 3 at $P$. If $\theta$ varies, then the locus of $\mathrm{P}$ is
a pair of straight lines
a circle
a straight line
None of these
Solution
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the point dividing the join of $\mathrm{A}$ and $\mathrm{B}$ in the ratio 2: 3 internally, then
$
\begin{array}{c}
\mathrm{x}=\frac{20 \cos \theta+15}{5}=4 \cos \theta+3 \Rightarrow \cos \theta=\frac{\mathrm{x}-3}{4} \\
\mathrm{y}=\frac{20 \sin \theta+0}{5}=4 \sin \theta \Rightarrow \sin \theta=\frac{\mathrm{y}}{4} \ldots \text { (ii) }
\end{array}
$
Squaring and adding (i) and (ii), we get the
required locus $(x-3)^{2}+y^{2}=16,$ which is
a circle