= Solution
Use the <complex potential of a two-dimensional potential flow> convention $W=\phi+i\psi$, for which $W'(z)=u-iv$. A positive <circulation> is counterclockwise. The two <point vortices> and the <two-dimensional point sink> give
$$
\boxed{W(z,t)=\frac{\Gamma}{2\pi i}\bigl[\log(z-c(t))+\log(z+c(t))\bigr]-\frac Q{2\pi}\log z.}
$$
Branch choices change additive constants, not the local <velocity> <field>. A vortex is transported by the nonsingular <velocity> due to the other vortex and the sink; its own singular term is omitted. At $z=c$ this gives
$$
\dot{\bar c}=\frac{\Gamma}{4\pi i c}-\frac Q{2\pi c},\qquad \dot c=\frac{-Q/(2\pi)+i\Gamma/(4\pi)}{\bar c}.
$$
The other vortex follows $-c$ by symmetry. Put $c=re^{i\theta}$, with $r(0)=a$ and $\theta(0)=0$. Multiplication by $\bar c$ gives
$$
r\dot r=-\frac Q{2\pi},\qquad r^2\dot\theta=\frac{\Gamma}{4\pi}.
$$
For $Q>0$, integration yields
$$
\boxed{r^2=a^2-\frac Q\pi t,\qquad\theta=-\frac{\Gamma}{4Q}\log\left(1-\frac{Qt}{\pi a^2}\right).}
$$
Equivalently, with $s=1-Qt/(\pi a^2)$,
$$
\boxed{c(t)=a\sqrt s\,e^{-i\Gamma\log s/(4Q)},\qquad z_1(t)=c(t),\quad z_2(t)=-c(t),\quad0\leq t<\frac{\pi a^2}{Q}.}
$$
The curves obey $\theta=(\Gamma/(2Q))\log(a/r)$, so they are <logarithmic spirals>. At $t=\pi a^2/Q$, both vortices reach the sink at the origin. For nonzero circulation they make infinitely many turns in that finite time; for zero circulation they approach radially. \b[The point-vortex evolution becomes singular at capture], so the formula does not define a regular continuation through the event.
\Image[/past-exam-of-the-options-i-course-of-the-university-of-oxford/2013-vortex-sink.png]
{title=Two equal vortices spiral symmetrically into a central sink and reach it at a finite capture time}
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