= Solution
Use $W=\phi+i\psi$ for the <complex potential of a two-dimensional potential flow>, so $W'(z)=u-iv$. The <Milne-Thomson circle theorem> says that if $f(z)$ is an incident <complex potential of a two-dimensional potential flow> whose singularities lie outside $|z|=a$, then
$$
W(z)=f(z)+\overline{f\!\left(\frac{a^2}{\bar z}\right)}
$$
makes $|z|=a$ a <streamline>. Indeed, on that circle $a^2/\bar z=z$, so $W=f+\bar f$ is real and the <stream function> is constant. This imposes the <no-penetration boundary condition>.
For the <two-dimensional potential doublet>, put $M=me^{i\alpha}$ and $f=M/(z-b)$, where $b>a$ is real. The theorem gives
$$
\boxed{W(z)=\frac{me^{i\alpha}}{z-b}
+\frac{me^{-i\alpha}z}{a^2-bz}}.
$$
Up to an additive constant, this is equivalently
$$
W(z)=\frac{me^{i\alpha}}{z-b}
-\frac{ma^2e^{-i\alpha}}{b^2(z-a^2/b)}.
$$
The <method of images> has placed the image <two-dimensional potential doublet> at $a^2/b$, inside the obstacle. We use the solution with zero added <circulation> and no imposed far-field flow; the original doublet is the only singularity in the liquid.
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