= Solution
Take positive vortex strength $\kappa$ to mean anticlockwise <circulation>. A <line vortex> at $z_0$ has local <velocity potential> $\kappa\arg(z-z_0)/(2\pi)$. The <image vortex at a plane wall> must have strength $-\kappa$ to enforce zero normal velocity at $y=0$. At the initial instant, a potential is therefore
$$
\boxed{\phi(x,y)=\frac{\kappa}{2\pi}\left[\arg((x-a)+i(y-b))-\arg((x-a)+i(y+b))\right].}
$$
This is a local, multivalued <velocity potential>: its <gradient> is single-valued away from the vortex, but the nonzero <circulation> precludes a globally single-valued potential around it. For a vortex centred at $(X,Y)$, the velocity is
$$
u_x=-\frac{\kappa}{2\pi}\frac{y-Y}{(x-X)^2+(y-Y)^2},\qquad
u_y=\frac{\kappa}{2\pi}\frac{x-X}{(x-X)^2+(y-Y)^2}.
$$
The real vortex and its opposite image give cancelling $u_y$ at the wall. A <point vortex> moves with the regular velocity induced by the other vortex, not its singular self-field. Evaluating the image field at $(X,b)$ gives $\dot X=\kappa/(4\pi b)$ and $\dot b=0$. Hence
$$
\boxed{(X(t),Y(t))=\left(a+\frac{\kappa t}{4\pi b},\,b\right).}
$$
The time-dependent <velocity potential> is obtained by replacing $a$ with $X(t)$ above; the image remains at $(X(t),-b)$. Reversing the convention for positive <circulation> reverses the translation direction.
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