The line joining (5,0) to ((10 θ, 10 θ) is divided internally in...

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.