= Solution
In a <circular orbit>, $r=a$ is constant, so the radial <acceleration> vanishes. The radial equation requires
$$
\boxed{h^2=ka^{n+3}}.
$$
For every $a>0$, this equation permits a real nonzero <specific angular momentum>, $h=\pm\sqrt{ka^{n+3}}$, and hence a <circular orbit>. This chooses $h$ for the desired radius; it does not claim that one prescribed value of $h$ permits every radius.
To test <circular-orbit stability for a power-law central force>, keep that chosen <specific angular momentum> fixed and write $r=a+\rho$, with $|\rho|\ll a$. The radial equation becomes
$$
\ddot\rho=\frac{h^2}{(a+\rho)^3}-k(a+\rho)^n.
$$
Expanding to first order and using $h^2=ka^{n+3}$ gives
$$
\ddot\rho=-\left(\frac{3h^2}{a^4}+kn a^{n-1}\right)\rho+O(\rho^2)
=-(n+3)ka^{n-1}\rho+O(\rho^2).
$$
For $n>-3$, the linearized equation is the <harmonic oscillator equation>, with
$$
\boxed{\Omega_r^2=(n+3)ka^{n-1}>0}.
$$
The radial disturbance undergoes bounded <small oscillations>, rather than growing. Equivalently, the <effective potential> has a strict local minimum: its second derivative at $a$ is $(n+3)ka^{n-1}>0$. By the <effective potential stability criterion>, sufficiently small radial disturbances remain small, proving \b[stability for $n>-3$]. For comparison, $n<-3$ gives exponential growth. At $n=-3$, the balancing value $h^2=k$ makes the radial acceleration identically zero, so a small radial velocity produces drift rather than a restoring oscillation.
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