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.
I think this problem as stated may have no solution ! I'm not completely sure about this but it's easier to post as an answer so here goes.
In the diagram below the points Y(-2.5,0), O(0,0) are as in your image (you have B instead of Y, A instead of O). The ellipse $p$ is a locus of arc endpoints from O of length 1, with the curvature varying (as complete circles they would be described by $x^2 + y^2 - 2ax = 0$, with $a$ varying - i.e. circles that touch the origin).
I don't have an analytic proof for this locus, but I found it in geogebra by drawing a conic through 5 points, each of which was such an arc, and then drew a few more such arcs just to check. Most of these aren't shown in the diagram, but they are there. Assuming this elliptical locus is correct, then your point C has to lie on the ellipse in order to have $\beta=1$.
The next observation is that the point $C$ has to be the nearest point to Y lying on $p$. This is because that is the only point on $p$ such that the tangent to $p$ at $C$ will be orthogonal to $YC$, a requirement if $YC$ is to be a tangent to the circle (i.e. $\beta$ must intersect $p$ orthogonally at $C$). I should note that I realize that this argument may be flawed: If orthogonality is not actually required and $YC$ can still be tangent to the circle arc at a point on $p$, there may still be a solution. One thought I had was, similarly to the locus of arc endpoints forming an ellipse, to look at the locus of tangents from $Y$ to the family of circles whose arcs form the possible $\beta$'s. My hunch is that that locus could not possibly be exactly identical to $p$ for a portion of $p$, and then deviate elsewhere (clearly it is not the whole ellipse).
What I found when carrying out this construction was that given the point $Y$ these conditions cannot be met: If $C$ is constructed as the point on $p$ nearest to $Y$, the point of tangency to the circle arc connecting to $C$ does not lie on $p$ for the point $Y$. What I found was that the point $H$ shown, and $F$ as the nearest point on $p$, does fit both requirements.
I'm not completely satisfied with this as an answer: the elliptical locus needs to be proven (maybe to work with the coordinates you generated for $C$ and to show that they describe an ellipse), and I suppose I'd have to say it's more of an increasingly strong hunch that there is no solution as stated than a proof! I'm wondering if the length of $\beta$ is allowed to vary, if then there is a solution?
Here is a drawing using Narasimham's numbers to precision of 5 -- with a 7 degree discrepancy in the angle. The grey and blue points close to each other correspond to the tangent at $107^{\circ}$ and the $\beta$ endpoint at $100^{\circ}$ (precision 5 in drawing), the latter obtained by using $\frac{\beta}{r}$ radians. Narasimham's formula looks right but I'd like to be able to demonstrate it and cannot.