The perimeter of the locus of the point $P$ which divides the line segment $\overrightarrow{Q A}$ internally…
- $8 \pi$
- $4 \pi$
- $\pi$
- $9 \pi$
Solution

Let $P \equiv(h, k)$
Also $P \equiv\left(\frac{4+6 \cos \theta}{3}, \frac{4+6 \sin \theta}{3}\right) \equiv(h, k)$
$\begin{aligned} & \therefore 3 h-4=6 \cos \theta \text { and } 3 k-4=6 \sin \theta \\ & \Rightarrow(3 h-4)^2+(3 k-4)^2=36 \\ & \Rightarrow\left(h-\frac{4}{3}\right)^2+\left(k-\frac{4}{3}\right)^2=4 \end{aligned}$ $\text { Perimeter }=2 \pi r=2 \pi \times 2=4 \pi \text {. }$
Asked in: AP EAMCET 2024 (21 May Shift 2)