The outer circle is $x^2+y^2=1$ and the smaller circle is $x^2+(y+1-r)^2=r^2$.
The arclength is parameterised anticlockwise with $s=0$ at the bottom as shown.
If we know $s_n$ and $s_{n+1}$ can we find the equation of the solid line and the point on the LHS where it intersects the smaller circle.
I would think that the point of intersection would be found after the finding the equation of the line between $s_n$ and $s_{n+1}$ and then substituting it into the equation for the smaller circle. Then you would take the negative $x$ solution.
Let $P_0$ be the point $s_0 = (0, –1)$ and $P_n$ the point on the circumference such that the minor arc $(P_n)(P_0) = s_n$. The point $P_{n+1}$ is similarly defined.
We further let the central angle subtended by the arc $(P_n)(P_0)$ be θ. Then $1.θ = s_n$, a known quantity.
Therefore, $P_n$ is located at $(–1.sin θ, –1.cos θ) = … = (–1.sin (s_n), –1.cos (s_n))$.
$P_{n+1}$ can similarly be located.
Apply two-point form to get L, the equation of the line $(P_n)(P_{n+1})$.
Solve L and C to get the required; where C is the small circle.
Note that the co-ordinates of $P_{n+1}$ = (+ve, -ve).