= Solution
{c}
Use the convention that a positive <vortex> has counterclockwise <circulation>, with <complex potential of a two-dimensional potential flow> $\Gamma\log(z-z_1)/(2\pi i)$. To make both coordinate axes <streamlines>, place images of strength $-\Gamma$ at $\bar z_1$ and $-\bar z_1$, and strength $+\Gamma$ at $-z_1$. The resulting potential is
$$
\boxed{W(z)=\frac{\Gamma}{2\pi i}
\log\frac{(z-z_1)(z+z_1)}{(z-\bar z_1)(z+\bar z_1)}.}
$$
Additive constants and consistent logarithm branches do not affect velocity. The <quotient> has unit modulus on either wall, so the <streamfunction> is constant there and there is no normal flow.
The <vortex> moves with the regular fluid velocity after removing its own singular <field>. Since $W'=u-iv$, evaluating the three image contributions at $z_1$ gives exactly
$$
\boxed{\frac{d\bar z_1}{dt}=-\frac{i\Gamma}{2\pi}
\left(\frac1{2z_1}-\frac1{z_1-\bar z_1}-\frac1{z_1+\bar z_1}\right).}
$$
The overall minus sign and the two reflected-term minus signs here agree with the source scan.
Write $z_1=x+iy$, with $x,y>0$. Taking real and imaginary parts yields
$$
\dot x=\frac{\Gamma x^2}{4\pi y(x^2+y^2)},\qquad
\dot y=-\frac{\Gamma y^2}{4\pi x(x^2+y^2)}.
$$
Dividing gives $dy/dx=-y^3/x^3$, which integrates to
$$
\boxed{\frac1{x^2}+\frac1{y^2}=\frac1{a^2}+\frac1{b^2}.}
$$
Put $\ell=ab/\sqrt{a^2+b^2}$. The first-quadrant branch has vertical and horizontal asymptotes $x=\ell$ and $y=\ell$. Its motion is to the right and downward: from large $y$ near the vertical asymptote toward large $x$ near the horizontal one. Both coordinates stay strictly positive, so no wall is crossed.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2006-options-quadrant-vortex.png]
{title=The three image vortices and the decreasing first-quadrant <vortex> path}
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