= Solution
The <effective potential> includes the angular part of the <kinetic energy>. Taking $U(0)=0$ gives $U=kr^4/4$, hence
$$
U_{\rm eff}(r)=\frac{J^2}{2mr^2}+\frac{k r^4}{4}.
$$
For $J\ne0$, this diverges at both zero and infinity and has a unique minimum. Finite-energy motion therefore remains between two radial <turning points>, or at the minimum itself. For $J=0$, the <centrifugal barrier> is absent, but $U\to\infty$ still bounds every finite-energy trajectory; radial motion can pass through the force centre.
A <circular orbit> requires $U_{\rm eff}'(r_c)=0$, so
$$
\boxed{r_c=\left(\frac{J^2}{mk}\right)^{1/6},\qquad
E_c=\frac34kr_c^4=\frac34 k^{1/3}\left(\frac{J^2}{m}\right)^{2/3}.}
$$
It is stable because $U_{\rm eff}''(r_c)=6kr_c^2>0$. These are nonzero-radius orbits for $J\ne0$; $J=0$ also permits the stationary particle at the origin. Boundedness does not mean that every noncircular orbit is closed.
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