Determining the geometry of an arc segment from two known points and a curvature Short version: How do i determine the profile of arc segment from two known points and curvature of the arc?
Long version: As an intro into modelling co-rotating twin-screw extruders i am trying to learn how the profiles of the screws are determined. For those of you who don't know, twin-screw extruders are machines which have two screws intermeshed in such a way that the action of a flight of one screw wipes the root of the other screw alternately. The below image shows the intermeshing of the screw (left) and the self-wiping action in a frontview perspective (right):


For this self-wiping action, the screw needs a special geometry determined by a special set of rules.
In the analysis i am following 
The geometry of self-cleaning twin-screw extruders by C. Rauwendaal.
The basic geometric properties of the barrel:


where $\alpha_i$ is the intermesh angle, $I_h$ is the intermesh height, $C_l$ is the centerline distance and finally $D$ and $D_r$ are the screw and root diameter respectively. For a given screw diameter $D$, these are related through some geometric considerations:
$$\frac{C_l}{D}=\cos\left(\frac{\alpha_i}{2}\right) \qquad
\frac{I_h}{D} = \frac{1}{2}\sin\left(\frac{1}{2}\alpha_i\right) \qquad
\frac{D_r}{D} = 2\frac{C_l}{D}-1$$
which basically implies that chosing an intermesh angle $\alpha_i$ fixes the centerline distance $C_l$, the intermesh height $I_h$ and the root diameter $D_r$.
Due to symmetry only a quarter of a screw needs to be examined. Such a quarter is further divisible into three sections:


*

*the tip -> $0<\alpha\leq\alpha_t/2$, 

*the flank -> $\alpha_t/2<\alpha\leq\alpha_t/2+\alpha_f$

*the root -> $\alpha_t/2+\alpha_f<\alpha\leq\alpha_t/2+\alpha_f+\alpha_r/2$


where $\alpha$ is the angle from the horizontal, $\alpha_t$ is the tip angle, $\alpha_f$ is the flank angle, $\alpha_r$ is the root angle. As the two screws have the same cross-sectional shape, the root angle should equal the tip angle. Furthermore, the sum of the tip angle, root angle and flank angle should be:
$$\frac{\alpha_r}{2}+\alpha_f+\frac{\alpha_t}{2}=\frac{\pi}{2}$$
Given the assumption $\alpha_r=\alpha_t$ as discussed above, this yields for the tip angle:
$$\alpha_t=\frac{\pi}{2}-\alpha_f$$
For the case of co-rotating screws, it is given that the flank angle has to be equal to the intermesh angle, i.e. $\alpha_f = \alpha_i$. From this follows then the tip angle $\alpha_t$ and the root angle $\alpha_r$.
Having determined all these geometric properties, I would now like to be able to reconstruct the profile of the screw. Taking the center of the screw as the origin $O$, the root and tip of the screw are easily described as they have constant radii with respect to $O$. However, the flank area with respect to $O$ has a variable radius which ranges from $\frac{D_r}{2}$ at the root angle to $\frac{D}{2}$ at the tip angle. In other words:
$$r = \begin{cases}
\frac{D}{2} & 0\leq\alpha<\frac{1}{2}\alpha_{t}\\
R\left(\alpha\right) & \frac{1}{2}\alpha_{t}\leq\alpha<\frac{1}{2}\alpha_{t}+\alpha_{f}\\
\frac{D_{r}}{2} & \frac{1}{2}\alpha_{t}+\alpha_{f}\leq\alpha<\frac{1}{2}\alpha_{t}+\alpha_{f}+\frac{1}{2}\alpha_{r}
\end{cases}$$
However, in the article it is given that the shape of the flank is determined by an arc of a circle (with a different unknown origin $C$) with radius $C_l$. 
What i would like to know is how I can determine the profile described by $R\left(\alpha\right)$ which is basically an arc segment between two points (i.e. $[r,\alpha]=[D/2, \alpha_t/2]$ and $[r,\alpha]=[D_r/2, \alpha_t/2+\alpha_f]$) with known curvature (radius $C_l$)?
Update: The answer of @ja27 gives a good and easy geometric method of constructing the arc in question:

However, as stated in the question (although maybe not clearly enough), I would like to be to describe the arc segment in terms of origin $O$ and angle $\alpha$ through function $R(\alpha)$, rather than origin $C$ and angle $\theta$.
 A: Take the two end points A and B as a line segment and find the perpendicular bisector. The center of the arc is going to be on this line. Now find the point of the line which is a distance $r$ (the radius) from the end point.



*

*You can find the distance $ m = \frac{1}{2} \| AB \| $ and then the distance $d = \sqrt{r^2-m^2}$ to find the point C.

*Or you can find the angle $\theta = \sin^{-1} \frac{m}{r}$ and then set $d = r \cos \theta$.
Edit 1
I think the follow-up question asks to find the red distance $r$ and angle $\theta$ below from the blue $R$ and $\varphi$:

These are found with a little trigonometry as
$$ \begin{aligned} 
  r \sin \theta & = R \sin \varphi \\
  r_O + r \cos\theta &= R \cos \varphi
\end{aligned} \Rightarrow
 \begin{aligned}
  r &= \sqrt{ R^2+r_O^2-2 R r_O \cos \varphi} \\
  \theta & = \tan^{-1} \left( \frac{R \sin \varphi}{R \cos\varphi -r_O} \right)
\end{aligned} $$
where $r_O$ is the radius of the outer circle. The arc radius is $R=r_O+r_I$ it turns out (I think).
