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

Accordingly, we are considering a subset of members of one specific inertial frame ("point particles sitting still in space relative to each other") whose distance relations among each other make them jointly having the shape of ... a circular disk (and its circular boundary).

$ A = \pi \, x^2 $

where $x$ apparently denotes the (length of the) radius of the circle to be considered; which, however, more usually is denoted by $r$.

If I move this circle with velocity $v$ in the x-direction,

Accordingly we're considering two distinct inertial frames (moving at constant speed $v$ straight along each other, in a direction which is in the plane of the circle):

• the inertial frame of which the constituents of the above-mentioned circle are a subset; let's call it "inertial frame $\mathsf D$" (as mnemonic for "domain"), and

• some specific other inertial frame ($\mathsf R$, for "range").

I would expect to see a contraction [...] in the x-direction and nothing on y-direction

For any two members of $\mathsf D$ (except the pairs oriented orthogonal to the direction of motion of $\mathsf R$'s members) their distance is larger than the length of their simultaneity projection in $\mathsf R$.

so [... the] area of a moving circle would be $A = \pi \, x^2 \, \sqrt{1 - \frac{v^2}{c^2}}$

But that's the area of each simultaneity projection in $\mathsf R$ of the circle! Instead, the area of the circle itself remains as before: $\pi \, x^2 $.

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

Accordingly, we are considering a subset of members of one specific inertial frame ("point particles sitting still in space relative to each other") whose distance relations among each other make them jointly having the shape of ... a circular disk (and its circular boundary).

$ A = \pi \, x^2 $

where $x$ apparently denotes the (length of the) radius of the circle to be considered; which, however, more usually is denoted by $r$.

If I move this circle with velocity $v$ in the x-direction,

Accordingly we're considering two distinct inertial frames (moving at constant speed $v$ straight along each other, in a direction which is in the plane of the circle):

• the inertial frame of which the constituents of the above-mentioned circle are a subset; let's call it "inertial frame $\mathsf D$" (as mnemonic for "domain"), and

• some specific other inertial frame ($\mathsf R$, for "range").

I would expect to see a contraction [...] in the x-direction and nothing on y-direction

For any two members of $\mathsf D$ (except the pairs oriented orthogonal to the direction of motion of $\mathsf R$'s members) their distance is larger than the length of their simultaneity projection in $\mathsf R$.

so [... the] area of a moving circle would be $A = \pi \, x^2 \, \sqrt{1 - \frac{v^2}{c^2}}$

But that's the area of each simultaneity projection in $\mathsf R$ of the circle!, which has the shape of an ellipse, with minor axis of length factor $\sqrt{1 - \frac{v^2}{c^2}}$ smaller than the diameter of the projected circle, oriented in the direction of motion of constituents of $\mathsf D$, and major axis of length equal to the diameter of the projected circle.

Instead, the area of the circle itself remains as before: $\pi \, x^2 $.