angle determined by tangent ratios I'm new here and not a math student so I apologize if this question is too dumb but I don't know where to begin to solve it.
Given the image below, I'm looking for the red "α" angle such that the ratio between the 2 red segments is equal to the ratio between the 2 blue segments. That's all you need to know. Every other angle is a known variable.

I'd lose my pride if someone told me the full answer but I need at least a starting point because my trigonometry skills are failing me.
 A: Assign names to points along the black tangent line: $A$ at the point of tangency, $R$ at the intersection with the red line, $B$ at the next intersection with a black line, $C$ at the intersection after that, and $D$ at the final intersection.
Also, let $O$ be the center of the circle.
I suppose the locations of the points $O$, $A$, $B$, $C$, $D$ are known.
Hence the angles $\theta = \angle AOB$ and $\beta = \angle BOC$ can be found and so can the ratio
$$ r = \frac{AB}{BD}. $$
The location of $R$ is initially unknown, and so is the angle
$\alpha = \angle ROB,$
but you know that the ratio of the red segments must equal the ratio of the blue segments.
The length of the lower red segment is $OA \times \tan\alpha$,
the total length of the two red segments is $OA \times \tan(\alpha + \beta)$,
and the length of the second red segment is the difference of those two lengths,
so the ratio of those two segments is
$$
\frac{OA \times \tan\alpha}{OA \times \tan(\alpha + \beta) - OA \times \tan\alpha}
= \frac{\tan\alpha}{\tan(\alpha + \beta) - \tan\alpha}.
$$
Since it is given that this ratio equals the ratio of the blue segments,
$$
\frac{\tan\alpha}{\tan(\alpha + \beta) - \tan\alpha} = r.
$$
Note that there is a formula (sometimes called the sum identity for tangent) that expresses $\tan(\alpha+\beta)$ in terms of $\tan\alpha$ and $\tan\beta$.
After using the sum identity there is not much more need for trigonometry.
That may be enough of a hint to solve the problem.
A: Thanks to David K's hint, I could carry on and find the solutions.
Below are my calculations (you have to read David's answer first). I decided to consider a slightly different "$r$" ratio to make it simpler: instead of the ratio between the small and the big segment, I used the small and the sum of the 2 segments. This means I did not have to add the $−tanα$ in the denominator of David's "$r$" fraction.

The quadratic equation brought to my attention the fact that there are actually 2 solutions to this geometrical problem.
This is the second one (I eyeballed the $α$ angle in this drawing, just like in the original one btw) :

So this one is the $\frac{-b+\sqrt{Δ}}{2a}$ root and the original one is $\frac{-b-\sqrt{Δ}}{2a}$
Why this way and not the other way around ? I haven't found proof other than the behavior of the software I use so far. Maybe I'll come back for a detailed answer on that later.
Careful : it may look like the red tangent is parallel to OD but that is pure coincidence.
