Can the radius of this constructed arc be found by knowing the two starting length values (1.6 and 3.7) and nothing else?
One expression with x and y as input (1.6 and 3.7) that gives me the result of the arc radius (2.196)?
arc.gh (7.2 KB)
I think the end-point of those two segments (1.6 and 3.7) define only two points in the plane, which are not enough to define a single circle (you need at least 3 points on circumference, or 1 point on circumference + center?)
a note: in your definition you have also set the direction of the Start Tangent point to be -X, that is an important costrain 
otherwise you get this indetermination:
so you actually have the center of the circle to be always along the Y axis
The circle is clearly defined with the command ARC SED, so this is not the problem. I want to calculate the radius using the length of the two lines I know using some trigonometric functions (SIN and COSIN).
I understand that
the point is, to have a SED arc you need 3 points and a tangent vector 
in case you are happy to have always the very same tangent vector as {-1,0,0} or {+1,0,0} (they both produce a circle with same diameter) despite x and y changing, then that is the equivalent of a circle per 3 points, where A and A’ are mirrored by the Y axis along B, because the center of the circle always falls on the same vertical as B
r = (x² + Y²) / 2Y
arc_Re.gh (14.8 KB)
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geometric solution.
relies on the fact that a triangle which contains a line that goes through the center and another point on the circle is always a right angle triangle
arc [geometric_solution].gh (15.3 KB)
Thank You very much!
Perfect!
Thank You Adel, nice solution too!
Beautiful!