Given a planar polygonal linkage defined by a sequence of $n$ hinge joints $(j_0,\,\cdots,\,j_{n-1},j_n = j_0)$ with links of fixed lengths $\lbrace\|j_{k+1}-j_k\|=d_k\ |\ 0\le k\lt n\rbrace$ between adjacent joints,
how can the set of radii be calculated that allow for placing all joints of the linkage on a circle with such a radius, while preserving the lengths of its links?
In view of the answers I want to stress that all joint shall lie on the same circle and, that the linkage is cyclically closed, i.e. its joints and links resemble a connected 2-regular graph.
There is already a question that is related to polygonal linkages: Is the area of a polygonal linkage maximized by having all vertices on a circle? which however was not concerned with determining the radius of the circle on which all joints lie when the area enclosed by the linkage is maximal.
Being able to calculate the smallest and the largest of those radii would be sufficient for my purposes, which I will state in a later edit.
In reply to the answers I'd like to add the following findings:
- a sharp lower bound for the radius on which all joints of a linkage can simultaneously lie, can be found by differentiating the radius of the circumcircle of a triangle w.r.t. one of its side-lengths:
$$\frac{\partial\left(\frac{x\,y\,z}{\sqrt{2(x^2y^2+y^2z^2+z^2x^2)-(x^4+y^4+z^4)}}\right)}{\partial x}=0\iff x=\sqrt{z^2-y^2} $$
which indicates that the circumcircle of a pair of adjacent links is minimal if the longer of the two is the hypotenuse of a right triangle and that the lower bound for entire linkage equals the maximum of all those minima. - if one divides the link-lengths by the by the sought radius, the resulting values are of the form $2\sin\left(\frac{\varphi_{ij}}{2}\right)$ and function for determining the radius can set up via the trigonometric addition theorems; that function can in turn be converted to rational one by utilizig $$\sin\left(\frac{\varphi_{ij}}{2}\right)=\frac{2t}{1+t^2}\quad \text{and}\quad \cos\left(\frac{\varphi_{ij}}{2}\right)=\frac{1-t^2}{1+t^2}$$