How does the area of moving circle change?

Your reasoning is correct. It is indeed possible and correct to calculate the are of some shape by using the contracted lengths.

For example, for a square with $A=xcdot y$ where in the square's rest frame $x=y$, we would calculate the new area (assuming the square moves in $x$ direction with velocity $v$) by

$$A'=y cdot x' = x^2{sqrt{1-frac{v^2}{c^2}}}=frac{A}{gamma}$$

(To avoid cunfusion: primed' indicates a measurement made on an object with relative velocity $v$ while non-primed is a measurement made in the object's rest frame)

In fact, it seems to me that the relationship $A'=frac{A}{gamma}$ also applies to volumes, for example for a cube with volume $V=xcdot ycdot z$ where in the cube's rest frame $x=y=z$, we have

$$V'=ycdot zcdot x'=yzxsqrt{1-frac{v^2}{c^2}}=frac{x^3}{gamma}=frac{V}{gamma}$$

The same would also apply to a sphere with $V=frac 4 3pi r^3$:

$$V'=frac 4 3pi r^2r'=frac 4 3pi r^2rsqrt{1-frac{v^2}{c^2}}=frac{V}{gamma}$$

It can thus be said, assuming constant velocity in one direction that

$$A'=frac{A}{gamma}tag{1}$$

$$V'=frac{V}{gamma}tag{2}$$

where $gamma$ is the Lorentz Factor $gamma = frac{1}{sqrt{1-frac{v^2}{c^2}}}$

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_{I am aware that the volume part isn't exactly related to the question, but I nevertheless wanted to expand a little in my answer, since the first part is basically just "yes"}