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The far-field derivative fixes the growing $n=1$ cosine coefficient to $U$ and removes all other growing modes. The possible logarithmic term has derivative $D_0/a$ at the wall, so the zero wall derivative removes it. For each decaying Fourier mode, the wall condition relates its coefficient to its growing partner: the $n=1$ cosine partner has coefficient $U{a}^2$, and every other partner vanishes. Thus the potential flow around a circular cylinder hasIndeed, ${\varphi }_r=U(1-a^2/r^2)\cos \theta$, which is zero at $r=a$ and tends to $U\cos \theta$ at infinity. The additive constant does not change the velocity. The equipotential curves satisfy $(r+a^2/r)\cos \theta =\text{constant}$. They approach vertical straight lines far away and meet the cylinder radially wherever the tangential velocity is nonzero. The two stagnation points are at $\theta =0,\pi$.

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges giveHere ${ϵ}_0$ is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let $d_1,d_2$ be the two distances. The equation $\Phi =0$ becomes $q_1{d}_2=-q_2{d}_1$, and henceFor a circle $r=\rho >0$ centred at the origin, its $\cos \theta$ coefficient must vanish: $q_1^2{r}_2=q_2^2{r}_1$. If $r_1,r_2>0$ and $r_1\ne r_2$, this givesSubstitute $q_2^2/q_1^2=r_2/r_1$ in the constant term:ThereforeThis is an Apollonius circle: the distance ratio $d_2/d_1$ is constant. The distinct-position assumption matters. For example, $r_1=r_2=1$, $q_2=-q_1$ gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.