= Solution
For
$$
V(r)=-\frac{km}{r-r_0},
$$
Binet's equation becomes
$$
\boxed{
u''+u=\frac{K}{(1-r_0u)^2}},
\qquad K=\frac{km^2}{\ell^2},\qquad 0<u<\frac1{r_0}.
$$
A circular orbit has constant $u=u_c$ and therefore satisfies
$$
K=u_c(1-r_0u_c)^2.
$$
The right side has maximum $4/(27r_0)$ at $u_c=1/(3r_0)$. Thus there are two circular orbits when $0<K<4/(27r_0)$, one marginal orbit at equality, and none above it.
Writing $u=u_c+\eta$ and linearizing gives
$$
\eta''+\frac{1-3r_0u_c}{1-r_0u_c}\eta=0.
$$
\b[Hence the outer orbit $u_c<1/(3r_0)$, equivalently $r_c>3r_0$, is stable; the inner orbit is unstable. These are the circular-orbit branches of the <shifted inverse-radius potential>.]
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