Answer
We can choose ${\bf{c}} = \left( {0,\frac{k}{J}e} \right)$, for some constant $e$, such that
${\bf{v}}\left( \theta \right) = \frac{k}{J}\left( { - \sin \theta ,\cos \theta } \right) + \frac{k}{J}\left( {0,e} \right)$
This equation implies that the velocity vector traces out a circle of radius $\frac{k}{J}$ with its center at the terminal point of ${\bf{c}}$ (as is shown in Figure 11).
Work Step by Step
The velocity vector of a planet as a function of the angle $\theta$ is given by Eq. (7):
${\bf{v}}\left( \theta \right) = \frac{k}{J}{{\bf{e}}_\theta } + {\bf{c}}$,
where ${{\bf{e}}_\theta } = \left( { - \sin \theta ,\cos \theta } \right)$ and ${\bf{c}}$ is an arbitrary constant vector.
Since ${\bf{c}}$ is an arbitrary constant vector, we can rotate our coordinate system in the plane of motion and write ${\bf{c}} = \left( {0,\frac{k}{J}e} \right)$, for some constant $e$. Thus,
${\bf{v}}\left( \theta \right) = \frac{k}{J}\left( { - \sin \theta ,\cos \theta } \right) + \frac{k}{J}\left( {0,e} \right)$
${\bf{v}}\left( \theta \right) = \frac{k}{J}\left( { - \sin \theta ,\cos \theta + e} \right)$
Notice that ${\bf{v}}\left( \theta \right)$ traces out a circle of radius $\frac{k}{J}$ centered at $\left( {0,\frac{k}{J}e} \right)$. Hence, as a planet revolves around the sun, its velocity vector traces out a circle of radius $\frac{k}{J}$ with its center at the terminal point of ${\bf{c}}$ (as is shown in Figure 11).
