= Solution
{c}
Assume $h\ne0$. The <velocity in polar coordinates> is $\dot{\mathbf r}=\dot r\mathbf e_r+r\dot\theta\mathbf e_\theta$. Hence
$$
(\dot{\mathbf r}\times\mathbf e_z)\cdot\mathbf r=r^2\dot\theta=h.
$$
Taking the <dot product> of the conserved <Laplace-Runge-Lenz vector> with $\mathbf r$ gives
$$
Ar\cos\gamma=m^2h^2-mkr.
$$
Consequently the <polar equation of a conic> is
$$
\boxed{r(\gamma)=\frac{m^2h^2}{mk+A\cos\gamma}=\frac{mh^2/k}{1+(A/(mk))\cos\gamma}.}
$$
Only <angles> for which the <denominator> is positive describe the physical trajectory. Its <eccentricity> is $A/(mk)$, giving an <ellipse>, <parabola>, or <hyperbola> when this ratio is respectively less than, equal to, or greater than $1$.
If $A=0$, the <angle> $\gamma$ is undefined, but the same <dot product> identity gives the <circular orbit> $\boxed{r=mh^2/k}$. If $h=0$, the trajectory is radial and the displayed division degenerates; for $k>0$, $\mathbf A=-mk\mathbf e_r$ and $\gamma=\pi$ on a fixed ray, so the identity does not determine $r$ as a function of a varying <angle>.
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