"In nonrelativistic limit of general relativity there is a correction to the Newtonian gravitational potential energy $−h/r^3$ with $h = αL^2/(mc)^2$, where c is the speed of light, α = GMm and L is the angular momentum"

Using this knowledge, I'm supposed to find the radius of circular orbits for a given m and L and decide which of them is stable. My question has to do with how I can actually determine the radii, but bear with me as I show my process thus far:

I e-mailed my professor and was told that I must SUBTRACT this correction factor from gravitational potential energy, which gives me:

$$ V(r) = \frac{-GMm}{r} - \frac{GMmL^2}{r^3(mc)^2}$$

I can find the effective potential to be:

$$V_{eff}(r) = \frac{L^2}{mr^3}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2} $$

Based on information in my textbook, I'd imagine that I must graph the effective potential and the straight line of my constant energy E, and the two radii will be the points where the energy line intersects the $V_{eff}(r)$ curve. My problem arises when I try to graph it.

If I draw a qualitative graph simply by using $\frac{1}{r^2} - \frac{1}{r} - \frac{1}{r^3}$ I get a curve that makes sense - I see a local maximum and a minimum, and I assume that the planet could stay "trapped" between the two "walls" of the minimum. When I attempt to use more "realistic" values, that is with a relatively large $x/r^3 $ term $(x = \frac{GMmL^2}{r^3(mc)^2})$, I get a curve with no apparent extrema that approaches 0 from the -y axis as r goes to infinity.

My question is (hopefully) a lot more general than just this example, $\textbf{"How do I find the potential radii of the orbiting planet using Kepler's Laws?"}$ Unless I've made a mistake in my process, I don't believe I can find them using this method. I'd imagine I could find them with a lot of calculus and rearranging, but I'm sure there must be a simpler way.