= Solution
{c}
For the <power law> <central potential> $V=\beta r^k$, differentiate the <effective potential>:
$$
V_{\mathrm{eff}}'(r)=\beta k r^{k-1}-\frac{J^2}{mr^3}.
$$
The <circular orbit> condition is $m\beta k r_0^{k+2}=J^2$. Thus, when $J\ne0$, $\beta k>0$ and $k\ne-2$,
$$
\boxed{r_0=\left(\frac{J^2}{m\beta k}\right)^{1/(k+2)}.}
$$
The sign condition $\beta k>0$ expresses attraction. At $k=0$ the <force> vanishes and there is no nontrivial <circular orbit>. At $k=-2$, attraction requires $\beta<0$ and the condition reduces to $J^2=-2m\beta$. In that exceptional case the <effective potential> is identically zero and every $r_0>0$ is a possible circular radius for the tuned $J$; there is no isolated preferred radius. A small radial <velocity> then gives $\ddot r=0$, rather than an oscillation.
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