I have two involutes of a circle. They are drawn from the same base circle, but one is offset from the other and faces the opposite direction (see plot below).
The parametric equations for the first involute are:
$$x = r_b (\cos(θ) + θ \sin(θ))$$ $$y = r_b (\sin(θ) - θ \cos(θ))$$
The parametric equations for the second involute are:
$$x = r_b (\cos(c-θ) - θ \sin(c-θ))$$ $$y = r_b (\sin(c-θ) + θ \cos(c-θ))$$
Where $r_b$ is the radius of the base circle, $c$ is a positive constant, and $θ \ge 0$ .
The angle their intersection point makes with the origin and $x$-axis is $c/2$ radians, but I can't find an expression that will give me either 1) the radius of the intersection point to complete the set of polar coordinates, or 2) the $(x,y)$ Cartesian coordinates.
I can't just set the parametric expressions equal to each other and solve this algebraically, because $θ\sin(θ)$ and $θ\cos(θ)$ are transcendental expressions and have no inverse.
I've tried putting in ranges of values for $c$, solving for the intersection numerically, and plotting the results, but the resulting points don't seem to line up with any of the functions - or combinations of functions - with which I am familiar.
The expression $θ_R = \sqrt{R² - r_b²} \div r_b$ gives the value of theta where the involute intersects a circle of radius $R$ (concentric with the base circle), from which I can get the coordinates of that intersection point, but that doesn't help when I don't have a value for the radius.
Plots
Involutes and base circle:
Locus of the point $x = c, y = \text{[radius of intersection]}$ with $r_b = 1$:
Same plot zoomed in to show detail with $c$ close to zero:
Solve the system
$$\begin{cases}\cos(\theta) + \theta\sin(\theta)=\cos(c-\phi) + \phi\sin(c-\phi), \\\sin(\theta)- \theta\cos(\theta)=\sin(c-\phi) - \phi\cos(c-\phi).\end{cases}$$
By squaring and adding the two equations,
$$1+\theta^2=1+\phi^2$$ and $\phi=\pm\theta$.
Then
$$1=\cos(\theta)(\cos(c-\phi) + \phi\sin(c-\phi))+\sin(\theta)(\sin(c-\phi) - \phi\cos(c-\phi)) \\=\cos(\phi-c-\theta)+\phi\sin(\phi-c-\theta).$$
If $\theta=\phi$,
$$\theta=\phi=\frac{\cos(c)-1}{\sin(c)}.$$
If $\theta=-\phi$, the equation is transcendental and must be solved numerically.
$$1=\cos(2\phi-c)+\phi\sin(2\phi-c).$$
Using
$$1=\sqrt{1+\phi^2}\cos\left(2\phi-c-\arctan\phi\right),$$
you can express $c$ as a function of $\phi$.