It suffices to show that any under the map $z \mapsto \frac{1}{z}$ any circle passing through the origin is mapped to a line. This follows from some basic geometric facts about the complex plane, but typically those aren't available when someone would encounter such a problem.
One circle passing through the origin the (radius $\tfrac{1}{2}$) circle parameterized by the polar equation $r_0(\theta) := \cos \theta$, and any other circle is given by dilating and rotating this circle about the origin, and so is given by
$$r(\theta) := A r_0(\theta - \theta_0) = A \cos (\theta - \theta_0).$$
Under the map $i: z \mapsto \frac{1}{z}$, this curve is mapped to the curve parameterized by
$$i(r(\theta)) = \frac{1}{A} \sec(\theta - \theta_0).$$ But this is precisely the polar parameterization of the line intersecting and perpendicular to the ray that (1) makes a (signed) angle $\theta_0$ with the $x$-axis and whose distance to the origin is $\frac{1}{A}$.