Here's the problem:
($2,5$ is $2.5$)
To determine $r$, I used Pythagoras and trigonometry to find that:
$\angle{BOC}=\dfrac{\beta}{r}$
$\tan{\dfrac{\beta}{r}}=-\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}$
As, from the graphic, $\angle{AOC}\in\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)$, when using $\arctan$, we get: $$\dfrac{\beta}{r}-\pi=-\arctan\left(\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}\right).\tag{1}$$
Plotting on WolframAlpha, an approximation of $r$ is $0.54$, which is what I get on Geogebra.
But I'm not satisfied. I relied on my eyes to know that the angle is $>\dfrac{\pi}{2}$, while if it belongs to $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, it's a different formula. It would be nice to be able to determine this for any given $\alpha$ and $\beta$ (where, of course, a circle like in the image would exist).
Do you know any other method? Like what I thought of is, if we manage to calculate one of the angles $\angle{ABC},\,\angle{AOC}$ or $\angle{ADC}$ (where $D\neq A$ is the other intersection of the line $(AB)$ and the circle), we'd be able to determine $r=\dfrac{\beta}{\angle{AOC}}$.
One other data that I got using tryhard analytic geometry: the coordonates of point $C$ are $\left(\dfrac{\alpha r}{r-\alpha},\dfrac{r\sqrt{(\alpha-r)^2-r^2}}{\alpha-r}\right)$.
Thank you in advance.
Edit
I just noticed that this always holds: $\angle{AOC}\in\left(\dfrac{\pi}{2},\pi\right)$. Thus $(1)$ always holds as long as $\beta$ is chosen in an adequate way.
So I guess here's the final result: given any construction as above, one has from $(1)$:
$$\beta=r\pi-r\arctan\left(\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}\right).$$
So, we look at $\beta$ as a continuous function of $r$ for now. Playing with Geogebra, I noticed that $\beta$ is increasing. If I'm not mistaken:
$$\beta'(r)=\pi-\arctan\left(\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}\right)+\dfrac{\alpha r}{\sqrt{(\alpha-r)^2-r^2}(\alpha-r)}$$
which is positive $(\arctan<\pi)$. This shows that $\beta$ is indeed increasing. Then we see that $$\lim_{r\to 0^+}\beta(r)=0$$ and $$\beta\left(\dfrac{\alpha}{2}\right)=\dfrac{\alpha\pi}{2}.$$
By the intermediate value theorem, this means that for any $\beta\in\left(0,\dfrac{\alpha\pi}{2}\right)$, we can make such a construction for exactly one possible $r$, which we can calculate numerically for specific examples.