= Solution
For an <incompressible flow>, $\nabla\cdot\mathbf u=0$. If the flow is also irrotational, then locally $\mathbf u=\nabla\phi$ for a <velocity potential>, and therefore
$$
\nabla^2\phi=\nabla\cdot\mathbf u=0.
$$
The boundary conditions are
$$
\left.\frac{\partial\phi}{\partial r}\right|_{r=a}=0,
\qquad
\nabla\phi\longrightarrow U\mathbf e_x\quad(r\to\infty),
\qquad
\int_0^{2\pi}\frac{\partial\phi}{\partial\theta}\,d\theta=\kappa.
$$
The fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the <potential flow around a circular cylinder with circulation>
$$
\boxed{\phi(r,\theta)=U\left(r+\frac{a^2}{r}\right)\cos\theta+\frac\kappa{2\pi}\theta}.
$$
Its velocity components are
$$
u_r=U\left(1-\frac{a^2}{r^2}\right)\cos\theta,
\qquad
u_\theta=-U\left(1+\frac{a^2}{r^2}\right)\sin\theta+\frac\kappa{2\pi r}.
$$
On $r=a$, $u_r=0$ and, with $\lambda=\kappa/(4\pi Ua)$,
$$
u_\theta=2U(\lambda-\sin\theta).
$$
The <Bernoulli equation> gives $p=p_\infty+\rho(U^2-u_\theta^2)/2$. The pressure force per unit length on the cylinder is
$$
\mathbf F=-a\int_0^{2\pi}p\,\mathbf e_r\,d\theta
=\frac{\rho a}{2}\int_0^{2\pi}u_\theta^2(\cos\theta\,\mathbf e_x+\sin\theta\,\mathbf e_y)\,d\theta.
$$
The $x$ component vanishes by symmetry, while $\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$ gives
$$
\boxed{\mathbf F=-4\pi\rho aU^2\lambda\,\mathbf e_y=-\rho\kappa U\,\mathbf e_y},
$$
in agreement with the <Kutta–Joukowski theorem>.
A <stagnation point> satisfies $u_r=u_\theta=0$. On the cylinder this means $\sin\theta=\lambda$. Thus $|\lambda|<1$ gives two surface stagnation points, $|\lambda|=1$ gives one coincident surface point, and $|\lambda|>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos\theta=0$. Solving $u_\theta=0$ then gives one physical exterior root when $|\lambda|>1$:
$$
\boxed{r=a\left(|\lambda|+\sqrt{\lambda^2-1}\right),
\qquad
\theta=\begin{cases}\pi/2,&\lambda>1,\\3\pi/2,&\lambda<-1.
\end{cases}}
$$
At $|\lambda|=1$ this root lies on $r=a$ and agrees with the single surface point.
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