Intersection of circle involutes

Question

I'm trying to find the intersection points between two spirals.

If we have one spiral as:

x1 = r*cos(θ) + r*θ*sin(θ)
y1 = r*sin(θ) - r*θ*cos(θ)

and another one as

x2 = -r*cos(θ) - r*θ*sin(θ) + r*π*cos(α)
y2 = -r*sin(θ) + r*θ*cos(θ) + r*π*sin(α)

The idea here is that by changing the value of α, the second spiral will rotate. For a specificed alpha value, is it possible to determine where the two spirals intersect?

For example, with r = 1, α = 0 and θ goes from 0 to 6*pi, as seen on the following equations and figure

x1 = cos(θ) + θ*sin(θ)
y1 = sin(θ) - θ*cos(θ)
x2 = -cos(θ) - θ*sin(θ) + π
y2 = -sin(θ) + θ*cos(θ)

Example of two spirals with highlighted intersections

Answer 1

Working in the assumption that $\alpha$ Is an arbitrary constant, we notice that $$x_2=-x_1+\pi r\cos(\alpha)$$ and $$y_2=-y_1+\pi r\sin(\alpha)$$

The functions intersect when $$x=x_1=x_2\text{ and } y=y_1=y_2$$ $$\Rightarrow 2x=\pi r\cos(\alpha)\text{ and } 2y=\pi r\sin(\alpha)$$ $$\Rightarrow x=\frac{\pi}{2} r\cos(\alpha)\text{ and } y=\frac{\pi}{2} r\sin(\alpha)$$ So for any $\alpha$, if the two spirals intersect, they will always intersect at $$\left(\frac{\pi}{2} r\cos(\alpha),\frac{\pi}{2} r\sin(\alpha)\right)$$

And now the question remains, given an $\alpha$, can we always find a $\theta$ such that $$\begin{cases} \sin(\theta)+\cos(\theta)=\frac{\pi}{2}\cos(\alpha)\\ \sin(\theta)-\cos(\theta)=\frac{\pi}{2}\sin(\alpha) \end{cases}$$ $$\Rightarrow\begin{cases} \sin(\theta)=\frac{\pi}{4}\sin(\alpha)+\frac{\pi}{4}\cos(\alpha)\\ \cos(\theta)=-\frac{\pi}{4}\sin(\alpha)+\frac{\pi}{4}\cos(\alpha) \end{cases}$$

The answer is not always Since for $x\in [0,2\pi)$, studying the two functions $$\begin{cases} f(x)=\frac{\pi}{4}\sin(x)+\frac{\pi}{4}\cos(x)\\ g(x)=-\frac{\pi}{4}\sin(x)+\frac{\pi}{4}\cos(x) \end{cases}$$ We notice that

$f(x)$ has a maximum at $x=\frac{\pi}{4}$ where $f(\frac{\pi}{4})\approx 1.11$ and a minimum at $x=\pi+\frac{\pi}{4}$ where $f(\pi+\frac{\pi}{4})\approx -1.11$

and

$g(x)$ has a maximum at $x=-\frac{\pi}{4}$ where $f(-\frac{\pi}{4})\approx 1.11$ and a minimum at $x=\pi-\frac{\pi}{4}$ where $f(\pi-\frac{\pi}{4})\approx -1.11$

So the main question becomes for which intervals do I get $$\begin{cases} -1\le f(x)\le 1\\ -1\le g(x)\le 1 \end{cases}$$ solving the equations $$f(x)=-1, f(x)=1, g(x)=-1,\text{ and } g(x)=1$$ We deduce the intervals which are too long if a calculation to write here.