 Solution

At an interior stagnation point, ${\psi }_r={\psi }_{\theta }=0$. The first equation isSince $\sin \theta >0$ inside the fluid, $r=2\sqrt{2}a\sin \theta$. AlsoOn the preceding locus the bracket is nonzero, so $\cos \theta =0$. ThereforeFor $\Omega >0$, $\psi <0$ in the interior, and its nested nonzero level curves are closed recirculating streamlines around this unique stagnation point.