 Solution

The Milne-Thomson circle theorem replaces a potential $f$ analytic inside the circle of radius $a$ by $f(z)+\overline{f(a^2/\bar{z})}$ outside it, when its original singularities are outside. The circle becomes a streamline, the imposed flow at infinity is preserved, and no extra circulation is introduced.

Use the convention that positive circulation is counterclockwise, so a vortex potential is $\Gamma \log (z-z_1)/(2\pi i)$ and its derivative is $u-iv$. With $b=a^2/{\bar{z}}_1$, the circle construction givesup to an additive constant and local logarithm branches. There is an image vortex of strength $-\Gamma$ at $b$ and a central vortex of strength $\Gamma$. Their zero net interior circulation is needed for the specified motion equation.

The original vortex moves with the regular velocity of the image flow, excluding its own singular velocity. ThusThe conjugation in this equation matches the printed source. Write $z_1=r{e}^{i\theta }$ and cancel $e^{-i\theta }$ to obtainReal and imaginary parts giveThe vortex moves uniformly on the circle determined by its initial radius, clockwise for positive $\Gamma$ in this convention.