Let $\text{circ}(x, y; R, c)$ be the characteristic function on the plane of a disk of radius $R$ and centre $c=(c_1,c_2)$:
$$\text{circ}\left(\frac{\sqrt{(x_1-c_1)^2+(x_2-c_2)^2}}{R}\right)=\begin{cases}1 & \text{if }\sqrt{(x_1-c_1)^2+(x_2-c_2)^2}< R, \\ \frac 12 & \text{if }\sqrt{(x_1-c_1)^2+(x_2-c_2)^2}=1, \\ 0 & \text{otherwise.} \end{cases} $$
In the case where $c=(0,0)$, the Fourier transform (with phase $-2\pi$ and unit normalization factor) of this function is \begin{equation} \mathcal F_r[\rho=\sqrt{f_1^2+f_2^2}]\left(\text{circ}\left(\frac{r}{R}\right)\right)=A\frac{J_1(2\pi R\rho)}{\pi R \rho} \end{equation} for $A=\pi R^2$, where $J_1$ is the Bessel function of the first kind. I'm looking for the transformation in the general case $c\ne 0$. I think this is related to the one-dimensional shift theorem, applied to both coordinates. In this case, I would have \begin{equation} \mathcal F_r[\rho]\left(\text{circ}\left(\frac{\sqrt{(x_1-c_1)^2+(x_2-c_2)^2}}{R}\right)\right)=A\exp\left(-2\pi i(c_1f_1+c_2f_2)\right)\frac{J_1(2\pi R\rho)}{\pi R \rho} \end{equation} for $\rho$ always equal to $\rho=\sqrt{f_1^2+f_2^2}$. Is this correct?