> _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"_](http://www.scholarpedia.org/article/Special_relativity:_kinematics#Galilean_and_Lorentz_transformations)) whose distance relations among each other make them jointly having the shape of ... a [circular disk (and its circular boundary)](https://en.wikipedia.org/wiki/Area_of_a_circle). > $ 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 $.