(Assume that the black line is tangent to the small circle passing through the point of contact of the 2 inner semicircles.)
This was forwarded to me by maths teacher. I tried solving the problem by using coordinates but did not succeed. Is it even possible to solve this using coordinates?
Also is there a geometrical way to solve this? I tried different things by using properties of tangents and angles in a semicircle. But again, I was not able proceed in the direction of the solution.
PS : @Blue has also discussed the otheer cases in the comments section, but the main assumption is the correct one as @Yves has pointed out in his answer.
Assuming the radius of a large half-circle to be $4$, in the triangle formed by its center, its contact point with the axis of symmetry and the center of the small circle, of radius $r$, we have by Pythagoras
$$(4-r)^2+4^2=(4+r)^2$$ and $r=1$.
Now in the triangle formed by the contact point of the large half-circle, the center of the small and the tangency point of the line, the hypothenuses is $3$ and a side is $1$, hence the other side $\sqrt8$. By similarity,
$$\frac y4=\frac1{\sqrt8}$$ while $$x=4-y.$$
This solution assumes that the line is the tangent to the small circle by the point of contact of the two large ones.
Update:
If we instead assume the line to be through the tangency point of the large and small circle, by similarity of the triangle $3-4-5$, we know the position of this tangency point, and again by similarity
$$\frac y4=\frac{\frac45}{3-\frac35},$$
giving