= Solution
Choose zero <electrostatic potential> at infinity. By <Coulomb's law> and the <superposition principle>, the two <point charges> give
$$
\boxed{\Phi(r,\theta)=\frac1{4\pi\epsilon_0}\left[\frac{q_1}{\sqrt{r^2+r_1^2-2rr_1\cos\theta}}+\frac{q_2}{\sqrt{r^2+r_2^2-2rr_2\cos\theta}}\right].}
$$
Here $\epsilon_0$ is the <vacuum permittivity>. For a nondegenerate zero-<equipotential curve>, the <electric charges> must have opposite signs. Let $d_1,d_2$ be the two distances. The equation $\Phi=0$ becomes $q_1d_2=-q_2d_1$, and hence
$$
q_1^2(r^2+r_2^2-2rr_2\cos\theta)=q_2^2(r^2+r_1^2-2rr_1\cos\theta).
$$
For a circle $r=\rho>0$ centred at the origin, its $\cos\theta$ coefficient must vanish: $q_1^2r_2=q_2^2r_1$. If $r_1,r_2>0$ and $r_1\ne r_2$, this gives
$$
\boxed{q_2=-q_1\sqrt{r_2/r_1}.}
$$
Substitute $q_2^2/q_1^2=r_2/r_1$ in the constant term:
$$
r_1\rho^2+r_1r_2^2=r_2\rho^2+r_2r_1^2,
\qquad (r_1-r_2)(\rho^2-r_1r_2)=0.
$$
Therefore
$$
\boxed{\rho=\sqrt{r_1r_2}.}
$$
This is an <Apollonius circle>: the distance ratio $d_2/d_1$ is constant. The distinct-position assumption matters. For example, $r_1=r_2=1$, $q_2=-q_1$ gives zero <electrostatic potential> everywhere away from the coincident <point charges>, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.
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