= Solution
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 point
$$
\boxed{r=2a\sin\theta.}
$$
At this boundary the outward radial vector from the cylinder center is $(\sin2\theta,-\cos2\theta)$, so the counterclockwise unit tangent is $\mathbf t=(\cos2\theta,\sin2\theta)$. Since $\mathbf e_r\mathbin\cdot\mathbf t=\cos\theta$ and $\mathbf e_\theta\mathbin\cdot\mathbf t=\sin\theta$, the <no-slip boundary condition> gives
$$
\boxed{u_r\cos\theta+u_\theta\sin\theta=a\Omega.}
$$
For 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 $\boxed{\psi=0}$.
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