Given the image below, how would you work out the position and radius of R?
Arc between two circles:
[Math] Calculate the arc between two circles
circles
Related Solutions
As others have mentioned in comments, your control points cannot be independent. Nevertheless, if we assume that a given configuration has the properties you want, we can analyze the geometry.
I'll consider the orange arc, $BE$, and I'll assume that both circles $A$ and $C$ overlap the interior of the orange circle, which I'll further assume has not degenerated into a line.
Let $a = |AB|$, $c = |CE|$, and $x=|AC|$; all of these can be considered known quantities. Let the (unknown) radius of the orange circle be $r = |PB| = |PE|$, where $P$ is the circle's center. Because radii $AB$ and $PB$ are perpendicular to a common tangent line at $B$, these segments lie on the same line; likewise for $CE$ and $PE$; consequently, $P$ lies at the intersection of the two extended radii $AB$ and $CE$, so that the angle $BPE$ is congruent to the angle between the vectors $AB$ and $CE$. Call the measure of that angle $\theta$; it, too, can be considered a known quantity.
Now, triangle $APC$ has sides of length $x$, $r-a$, and $r-c$ (the last two because of the assumed overlap of circles), with angle $\theta$ between the last two. By the Law of Cosines:
$$x^2 = (r-a)^2 + (r-c)^2 - 2 (r-a)(r-c) \cos\theta$$
Solve this quadratic equation for $r$, and you can calculate whatever else you need to know: arc length, location of $P$, equation of the orange circle, etc.
(The equation of the orange circle can be expressed in a form that degenerates into a line equation as $r$ approaches infinity. Note that, in such a degenerate case, $\theta = 0$.)
You already have one good answer, but here are some pointers to a more pedestrian method easily accessible to a student of trigonometry.
Given the Cartesian coordinates $(x_0,y_0)$ of the center of a circle in a Cartesian plane, and the coordinates $(x_1,y_1)$ and $(x_2,y_2)$ of two points on that circle, you first want to find the direction from the circle's center to each point.
In many computer programming environments you can do this with an atan2 function;
as already explained in an answer to another question,
the formula for the first of the points is
$$ \theta_1 = \mbox{atan2}(y_1 - y_0, x_1 - x_0).$$
Alternatively, using more traditional trigonometry you can write
$$ \theta_1 = \begin{cases} \tan^{-1} \left(\frac{y_1 - y_0}{x_1 - x_0}\right) & \mbox{if } x_1 - x_0 > 0, \\ \tan^{-1} \left(\frac{y_1 - y_0}{x_1 - x_0}\right) + \pi & \mbox{if } x_1 - x_0 < 0, \\ \frac\pi2 & \mbox{if } x_1 - x_0 = 0 \mbox{ and } y_1 - y_0 > 0, \\ -\frac\pi2 & \mbox{if } x_1 - x_0 = 0 \mbox{ and } y_1 - y_0 < 0, \\ \end{cases} $$
The reason this is so complicated is that $\tan^{-1}$ gives answers only in the range $(-\frac\pi2, \frac\pi2),$ which gives you the directions to points only on a semicircle, not the whole circle, so we need to do something to get the points on the other semicircle; and of course we cannot do anything at all with $\frac{y_1 - y_0}{x_1 - x_0}$ when the denominator of that ratio would be zero.
Either of these methods gives you an angle measured in the anti-clockwise sense from the direction of the positive $x$-axis to the direction from $(x_0,y_0)$ to $(x_1,y_1)$.
Once you have such angles for both points, $\theta_1$ and $\theta_2$, subtract the angle of the point you are coming from from the angle of the point you are going to, and this will give you the angle of the anti-clockwise arc from one point to the other ...
... with one small hitch, which is that the answer should be in the range $[0, 2\pi)$ since you don't need to go around the circle more than once to reach the desired point, but subtracting two angles from either of the methods above will not always be in that range.
So after computing the difference of the angles, $\theta$, if you don't have $0 \leq \theta < 2\pi$ you "normalize" the angle by adding or subtracting a whole multiple of $2\pi$ to get a result $\theta'$ that is in the range $[0, 2\pi)$.
Once you have a normalized angle $\theta'$ (measured in radians, of course!), the anti-clockwise arc length is $\theta' r$,
To get the clockwise arc length, subtract the anti-clockwise arc from $2\pi r.$
Best Answer
According to the given, there will be no solution. Imagine the case that if circle B is placed further apart from circle A, the new required arc length will definitely be longer than the original. Therefore, we have to know D, the distance between two centers.
The example you mentioned is a simple one because R, the radius of the circle-to-be-constructed, is known. In that case, we only need to draw the red arc (centered at A with radius = R + a) and similarly the green arc. Their intersection will yield the point C which is the center of the required circle. In this setting, “the triangle inequality” requires $(R + a) + [R + b] \ge D$. That is, $R \ge (D – a – b)/2$, as mentioned by @dxiv.
If R must be a calculated result, then according to the cosine rule (applied to $\triangle ABC$), one more angle must also be known. For example, if $\alpha$ is known, then
$$(R + a)^2 + D^2 – 2(R + a) D \cos \alpha – (R + b)^2 = 0$$
After simplification, we will get a quadratic equation in R. From which, R can then be found if the discriminant is non-negative.