Solution:
Consider two consecutive circles ωi and ωi+1. Let Qi,Qi′ be two points on ωi and Qi+1,Qi+1′ on ωi+1 such that Qi,Pi and Qi+1 are colinear and also Qi′,Pi and Qi+1′. Then QiQi′=2∠QiPiQi′=2∠Qi+1PiQi+1′=∠Qi+1Qi+1′. Refer to the center of ωi as Oi. The previous result shows that the lines OiQi and Oi+1Qi+1 meet at the same angle as the lines OiQi′ and Oi+1Qi+1′, call this angle ψi. ψi is a function solely of the circles ωi and ωi+1 and the distance between them (we have just showed that any two points Qi and Qi′ on ωi give the same value of ψi, so ψi can't depend on this.) Now, the geometry of ωi and ωi+1 is the same for every i, so ψi is simply a constant ψ which depends only on r. We know 6ψ=0mod2π because Q7=Q1.
We now compute ψ. It suffices to do the computation for some specific choice of Qi. Take Qi to be the intersection of OiOi+1 and ωi which is further from Oi+1. We are to compute the angle between OiQi and Oi+1Qi+1 which is the same as ∠OiOi+1Qi+1. Note the triangle △OiPiOi+1 is isosceles, call the base angle ξ. We have
∠OiOi+1Qi+1=∠OiOi+1Pi+∠PiOi+1Qi+1=ξ+(π−2∠Oi+1PiQi+1)=ξ+(π−2(π−∠QiOi+1Pi−∠PiQiOi+1))=ξ−π+2(ξ+(1/2)∠PiOiOi+1)=ξ−π+2(ξ+(1/2)ξ)=4ξ−π.
So we get 6(4ξ−π)=0mod2π. Noting that ξ must be acute, ξ=π/12,π/6,π/4,π/3 or 5π/12. r is uniquely determined as (1/2)secξ so there are 5 possible values of r.