If the area of an expanding circular region increases at a constant rate with respect to time, then the rate of increase of the perimeter with respect to the time

Varies inversely as radius

Let A be the area and P be the perimeter of the circular region of radius r. Then,

$A=\pi r^2$ and $P=2 \pi r$

$\Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}$ and $\frac{d P}{d t}=2 \pi \frac{d r}{d t}$

It is given that $\frac{d A}{d t}$ = k (constant).

$\Rightarrow \frac{d r}{d t}=\frac{k}{2 \pi r}$

∴  $\frac{d P}{d t}=2 \pi \times \frac{k}{2 \pi r}=\frac{k}{r} \propto \frac{1}{r}$