Does an airfoil in supersonic flow have zero circulation?
Maybe I'm missing something obvious to you, but why do you think there's zero circulation?

Take the simplest case of an infinitely thin 2D flat plate in linearized supersonic flow. The solution in the perturbed velocity potential ($\hat{\phi}$) is:

$$\hat{\phi}=
    \begin{cases}
    0 &amp; x\pm\beta z&lt;0 \\
    \frac{1}{\beta}\alpha(x-\beta z) &amp; 0\le x-\beta z\le c, &amp; z&gt;0 \text{ (upper surface)}\\
    -\frac{1}{\beta}\alpha(x+\beta z) &amp; 0\le x+\beta z\le c, &amp; z&lt;0 \text{ (lower surface)}\\
    \pm\frac{1}{\beta}(\alpha c) &amp; x\pm\beta z&gt;0
    \end{cases}
$$

where $c$ is the chord length of the flat plate, $x$ and $z$ are the horizontal and vertical coordinates, $\alpha$ is the angle of attack and $\beta=\sqrt{M_\infty^2-1}$ is the supersonic Prandtl-Glauert factor.

The velocity field ($\vec{V}$) can  be obtained as:

$$\vec{V} = \begin{bmatrix}V_\infty \\ 0\end{bmatrix}+V_\infty\nabla{\hat{\phi}} = 
\begin{bmatrix}V_\infty\pm\frac{1}{\beta}\alpha V_\infty \\ -\alpha V_\infty\end{bmatrix}$$

where $V_\infty$ is the free-stream airspeed. The first equation is plus for upper surface and minus for lower surface.

Take an infinitely thin closed path ($C$) that wrap around the upper and lower surfaces, and calculate the circulation ($\Gamma$):

$$\Gamma = \oint_C\vec{V} \cdot \hat{n} ds=\frac{2\alpha c V_\infty}{\beta}$$

This is obviously non-zero. In fact, this can be readily seen by the potential jump in the wake.

And when we calculate the lift coefficient ($C_l$) of this flat airfoil using the Kutta-Joukowski Theorem ($L'=\rho_\infty V_\infty \Gamma$), we retrieve the classic supersonic lift relation:

$$C_l=\frac{L'}{1/2\rho_\infty V_\infty^2 c}=\frac{4\alpha}{\beta}$$