= Solution
For the attractive <central force> $\mathbf F=-k\mathbf r$, with $k>0$, the <potential energy> is $U(r)=kr^2/2$ up to an additive constant. Conservation of <angular momentum> confines the motion to a plane; writing its magnitude as $J=mr^2\dot\varphi$, the <kinetic energy> is
$$
T=\frac12m\dot r^2+\frac12mr^2\dot\varphi^2=\frac12m\dot r^2+\frac{J^2}{2mr^2}.
$$
At fixed $J$, the radial <energy> equation becomes $E=m\dot r^2/2+U_{\rm eff}(r)$, where
$$
\boxed{U_{\rm eff}(r)=\frac{J^2}{2mr^2}+\frac12kr^2.}
$$
The <effective potential> combines the true <potential energy> with the angular-motion contribution. Its negative derivative gives the radial equation $m\ddot r=-U_{\rm eff}'=J^2/(mr^3)-kr$; the first term is the centrifugal contribution arising from elimination of the angular coordinate.
A <circular orbit> is a fixed radial position at a stationary point:
$$
\boxed{r_0=\left(\frac{J^2}{mk}\right)^{1/4},\qquad
\Omega=\frac J{mr_0^2}=\sqrt{\frac km}.}
$$
For $J\ne0$, $U_{\rm eff}''(r_0)=4k>0$, so this is a stable minimum. Small radial disturbances have <angular frequency> $2\sqrt{k/m}$. For $J=0$, there is no nonzero-radius <circular orbit>.
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