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ForBinet's equation becomesA circular orbit has constant $u=u_c$ and therefore satisfiesThe 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 givesHence 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.