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 

Let $A$be the area and $P$be the perimeter of the

circular region of radius $r$ Then,

$A=\pi r^2$ and $P=2\pi r$

$\Rightarrow \frac{dA}{dt}=2\pi r\frac{dr}{dt}$ and $\frac{dP}{dt}=2\pi \frac{dr}{dt}$

It is given that $\frac{dA}{dt}=k\left(\text{ constant }\right)$

$\begin{array}{l}\Rightarrow \frac{dr}{dt}=\frac{k}{2\pi r}\\ \therefore \frac{dP}{dt}=2\pi \times \frac{k}{2\pi r}=\frac{k}{r}\propto \frac{1}{r}\end{array}$