= Solution
At an interior <stagnation point>, $\psi_r=\psi_\theta=0$. The first equation is
$$
\psi_r=a\Omega\sin\theta
\left(1-\frac{8a^2\sin^2\theta}{r^2}\right)=0.
$$
Since $\sin\theta>0$ inside the fluid, $r=2\sqrt2a\sin\theta$. Also
$$
\psi_\theta=a\Omega\cos\theta
\left(r-12a\sin\theta+\frac{24a^2\sin^2\theta}{r}\right).
$$
On the preceding locus the bracket is nonzero, so $\cos\theta=0$. Therefore
$$
\boxed{(x,y)=(0,2\sqrt2a).}
$$
For $\Omega>0$, $\psi<0$ in the interior, and its nested nonzero level curves are closed recirculating streamlines around this unique stagnation point.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1-touching-cylinder-streamlines.png]
{title=<Streamlines> around two touching cylinders}
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