We know that the area of a circle at rest frame is

$$A=\pi x^2$$ from $\pi xy$ where y=x

If I move this circle with velocity $v$ in the x-direction, I would expect to see a contraction (I'm not talking about optical observation since in that case it will still look like a circle) in the x-direction and nothing on y-direction so my area will be

$$A=\pi x'y$$

Since length in y-direction hasn't changed and $x'=x{\sqrt{1-\frac{v^2}{c^2}}}$, area of a moving circle would be

$$A=\pi x^2 \sqrt{1-\frac{v^2}{c^2}}$$

I couldn't find anything on internet about contractions in area, this above reasoning would be how I would attack the problem, is that correct?