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The inner cylinder is $x^2+(y-a{)}^2=a^2$. With $x=r\cos \theta$ and $y=r\sin \theta$, this becomes $r^2-2ar\sin \theta =0$, so away from the touching pointAt this boundary the outward radial vector from the cylinder center is $(\sin 2\theta ,-\cos 2\theta )$, so the counterclockwise unit tangent is $\mathbf{t}=(\cos 2\theta ,\sin 2\theta )$. Since ${\mathbf{e}}_r\cdot \mathbf{t}=\cos \theta$ and ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=\sin \theta$, the no-slip boundary condition givesFor a streamfunction in polar coordinates, no penetration through a stationary material curve makes $\psi$ constant along that curve. Both cylinders touch at the origin, so continuity forces their constants to agree. Since adding a constant does not alter velocity, choose their common value as $\boxed{\psi =0}$.

In the exterior domain $r>a$, the streamfunction obeysUsing the Streamfunction in polar coordinates conventionthe no-slip condition on the stationary disk isThus $\psi$ is constant on the disk and ${\psi }_r=0$ there. Choose the arbitrary constant to be zero:

The imposed shear ${\mathbf{u}}_{\infty }=\gamma y{\mathbf{e}}_x$ has streamfunctionThe far-field condition is therefore

A stagnation point of a fluid flow is a point at which the velocity field vanishes. For a two-dimensional streamfunction, it is a critical point satisfying both first spatial derivatives of the streamfunction equal to zero.