 Solution

Use $W=\varphi +i\psi$ for the complex potential of a two-dimensional potential flow, so $W^{\prime }(z)=u-iv$. 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 $|z|=a$, thenmakes $|z|=a$ 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 $M=m{e}^{i\alpha }$ and $f=M/(z-b)$, where $b>a$ is real. The theorem givesUp to an additive constant, this is equivalentlyThe 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.