What is the Fourier transform for the given two dimensional airy function,
$$f(x,y) = \frac{J_1(r)}{r}\,.$$
Where $J_1$ is the Bessel function of the first kind, order one. And $r=\sqrt{x^2+y^2}$.
Written explicitly,
$$\mathcal{F}\left[f(x,y)\right] = \int^\infty_{-\infty}\int^\infty_{-\infty}f(x,y)\exp\left[ -i2\pi\left( f_xx+f_yy\right)\right]\, dx\,dy\, .$$
A radial function has a radial Fourier transform, and as stated by reuns in the comments we just need to evaluate
$$\int_{0}^{+\infty}J_1(r)\int_{0}^{2\pi} e^{-2\pi i r \cos(t) w}\,dt\,dr $$ or $$ 2\pi \int_{0}^{+\infty} J_1(r) J_0(2\pi r w)\,dr. $$ Using the Fourier transforms of $J_0$ and $J_1$, the last integral equals zero if $2\pi w > 1$, $\pi$ if $2\pi w=1$ and $2\pi$ if $0\leq 2\pi w<1$. In other terms the Fourier transform of your function is a multiple of the indicator function of the unit circle, halved on the boundary.