The velocity of any point PP on a wheel can be written as the sum of two velocities: the velocity V→V→ of the center OO and the velocity ω→×OP→ω→×OP→ of rotation about the center, where ω→ω→ is angular velocity (perpendicular to the plane of the wheel).
A wheel turns without sliding with respect to a given path if the velocity of the contact point between wheel and path vanishes. Let then CC and C′ be the contact points of the two wheels. We have
If v→C=0 then V→=−ω→×OC→ and This cannot vanish, unless C=C′. So the assumption that both circles turn without sliding is false.
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Ah, that makes it definitely less pleasant. Your previous version was a quadratic equation, easy to solve. Now you get a third degree equation. This can be solved, but doing that is as pleasant as a third degree interrogation. You'll find many methods on the Internet, but whatever method you use, the law of conservation of misery applies.
Perhaps the easiest solution in practice is to use a numerical method like Newton-Raphson. It does also have its difficulties, but these are not too bad in comparison.
I haven't looked further at your equations, as I'm now busy with my own calculations, and I'm already confused enough...