= Solution
In a short time $dt$, the radius vector sweeps an oriented area $d\mathcal A=\tfrac12r^2d\theta$. Thus
$$
\boxed{\frac{d\mathcal A}{dt}=\frac h2.}
$$
The constant $h$ is twice the oriented <areal velocity>, or the <angular momentum> per unit mass. Its constancy is <Kepler's second law>: equal areas are swept in equal times.
For a <circular orbit> of radius $a$, set $\ddot r=0$ in the radial equation. The balance is
$$
h^2=GMa,\qquad |\dot\theta|=\frac{|h|}{a^2}=\sqrt{\frac{GM}{a^3}}.
$$
Hence a <circular orbit> exists for any $a>0$ with this nonzero $|h|$. Its period gives <Kepler's third law>:
$$
\boxed{T=2\pi\sqrt{\frac{a^3}{GM}},\qquad T^2=\frac{4\pi^2a^3}{GM}.}
$$
To check <circular-orbit stability for a power-law central force> directly, keep $h$ fixed and set $r=a+\rho$. Expanding
$$
\ddot r=\frac{h^2}{r^3}-\frac{GM}{r^2}
$$
about the balanced radius gives
$$
\ddot\rho=\left(-\frac{3h^2}{a^4}+\frac{2GM}{a^3}\right)\rho+O(\rho^2)
=-\frac{GM}{a^3}\rho+O(\rho^2).
$$
The linear radial perturbation therefore obeys
$$
\boxed{\ddot\rho+\Omega^2\rho=0,\qquad \Omega^2=GM/a^3>0.}
$$
It remains bounded and oscillates, proving stability under the stipulated small perturbations. Equivalently, the specific <effective potential> $h^2/(2r^2)-GM/r$ has a strict minimum at $a$.
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