I have looked at Goldberg's paper referenced by J. J. Castro in his excellent answer. It turns out that there is a simpler (and more general way) to generate Goldberg's non-circular solutions, so I thought that I would just mention that.
The idea is to consider the 'rotor' (Goldberg's term) as fixed and as the envelope of a circle in periodic motion. It's a bit easier if you use complex notation, i.e., think of $\mathbb{R}^2$ as $\mathbb{C}$. Consider a circle of radius $r>0$ and center $z_0=1$, parametrized by $z(s) = 1 + r\mathrm{e}^{is}$, and move it by a circle of rigid motions: $R_t(z) = \mathrm{e}^{it}z + a(t)$, where $a$ is periodic of period $2\pi/n$ for some integer $n\ge 3$. Now take the two envelopes of this $1$-parameter family of circles $R_t\bigl(z(s)\bigr)$. Because of the periodicity of $a$, these will also be envelopes of the circles of radius $r$ centered at the other $n$-th roots of unity, and hence these envelopes will have the property that, when they are moved by the inverse of $R_t$, they will remain tangent to the $n$ circles of radius $r$ centered on the $n$-th roots of unity.
A simple computation shows that the two envelopes of this family are given by
$$
E_\pm(t) = \mathrm{e}^{it}+a(t) \pm \frac{r\bigl(\mathrm{e}^{it}-ia'(t)\bigr)}{\bigl|\mathrm{e}^{it}-ia'(t)\bigr|}.
$$
(Note that this is well-defined as long as $\mathrm{e}^{it}-ia'(t)$ never vanishes, which can be ensured, for example, by choosing $a$ so that $|a'(t)|<1$.)
Setting $a=0$ (or a constant) gives the trivial circle solution. Below, I give an example of $E_-$ (the interior envelope) with $n=3$, $r=1/2$, and $a(t) = \tfrac17\sin(3t)$, or, more precisely, $R_{-s}(E_-)$ as $s$ varies between $0$ and $2\pi$. (Replacing this by $a(t) = \tfrac16\sin(3t)$ yields a nonconvex solution. In general, when $a$ is allowed to be large, the envelopes will have cusps.)