If the area of a circle increases at a uniform rate,

We know that the area of circle, A = πr2, where r = radius of the circle

And, perimeter = 2πr

According to the question, we have

$\frac{d \mathrm{~A}}{d t}=\mathrm{K}$, where $\mathrm{K}=$ constant

$\Rightarrow \frac{d}{d t}\left(\pi r^{2}\right)=\mathrm{K} \Rightarrow \pi \cdot 2 r \cdot \frac{d r}{d t}=\mathrm{K}$

So, $\frac{d r}{d t}=\frac{\mathrm{K}}{2 \pi r}$

Now, perimeter $c=2 \pi r$

Differentiating w.r.t t, we get

$\frac{d c}{d t}=\frac{d}{d t}(2 \pi r) \quad \Rightarrow \frac{d c}{d t}=2 \pi \cdot \frac{d r}{d t}$

$\frac{d c}{d t}=2 \pi \cdot \frac{\mathrm{K}}{2 \pi r}=\frac{\mathrm{K}}{r}$ [From (1)]

$\frac{d c}{d t} \propto \frac{1}{r}$

Therefore, it’s seen that the perimeter of the circle varies inversely as the radius of the circle.