Here is another (hopefully better) attempt, but first tackle a simpler case as a demo of the workflow.

    p1 = Plot3D[HeavisidePi[1/2 (x^2 + y^2)], {x, -2, 2}, {y, -2, 2}
       , PlotStyle -> Darker@Cyan
       , ImageSize -> 300
       , AxesLabel -> Automatic
       , Exclusions -> All
       , ExclusionsStyle -> HatchShading[.4, Darker@Cyan]
       , SphericalRegion -> True
       , PlotRangePadding -> Scaled[.05]
       ];
    
    FourierTransform[
      FunctionExpand@HeavisidePi[1/2 (x^2 + y^2)], {x, 
       y}, {ω1, ω2}] // FunctionExpand

$\frac{I_1\left(\sqrt{-\text{$\omega $1}^2-\text{$\omega $2}^2}\right)}{\sqrt{-\text{$\omega $1}^2-\text{$\omega $2}^2}}$
    
    p2 = Plot3D[BesselI[1, Sqrt[-ω1^2 - ω2^2]]/
       Sqrt[-ω1^2 - ω2^2]
       , {ω1, -10, 10}
       , {ω2, -10, 10}
       , PlotStyle -> ColorData["HTML", "Aquamarine"]
       , PlotRange -> {-0.2, 0.6}
       , ImageSize -> 300
       , AxesLabel -> Automatic
       , PlotPoints -> 20
       , MaxRecursion -> 2
       , SphericalRegion -> True
       , PlotRangePadding -> Scaled[.05]
       ];
    
    Grid[{{p1, p2}}]

---

[![enter image description here][1]][1]

---

**Comments on the more general case**

Using a function with a disk of radius R and situated at an offset:

    f[x_, y_, c1_, c2_, R_] := 
     FunctionExpand@HeavisidePi[1/(2 R) ((x - c1)^2 + (y - c2)^2)]

---

While computing the Fourier transform, this evaluates instantly but with `c1=0` and `c2=0`. Note that the radius can be set differently.

    FourierTransform[
      f[x, y, 0, 0, 1], {x, y}, {ω1, ω2}] // FunctionExpand

A workaround (check!) can be to add an offset to `ω1` and `ω2` after the transform calculation in order to shift the center of the 3D sinc-like plot. As an example:

    FourierTransform[
       Evaluate@f[x, y, 0, 0, 3], {x, 
        y}, {ω1, ω2}] /. {ω1 -> ω1 - 
         1, ω2 -> ω2 - 1/2 } // FunctionExpand

I cannot comment on its math correctness so you can check or devise another strategy to evaluate the Fourier transform. I am not aware of such a strategy at this point. Perhaps the transform will evaluate after several minutes more and I haven't waited for more than five minutes on my machine before quitting the calculation.

  [1]: https://i.sstatic.net/aZWsz.png