One approach is to let the midpoint of the curve be a random point within (curve length)/2 of both of the starting point and the end point, and then iterate.  For curves of length 1 between (0,0) and (1/2,0) that gives results like this:

[![three random curves][1]][1] 

In more detail, to connect a starting point and ending point by a curve of length $2c$, draw circles of radius $c$ about both points, and randomly pick a point in the intersection to be the midpoint of the curve.  So from the points at distance 0 and 1 on the curve, calculate the point at distance 1/2, and then the points at distance 1/4 and 3/4, etc.  For the pictures above, I went down to distance 1/1024, picking points in the inscribed rhombus rather than the curved shape to simplify the algebra.

[![diagram][2]][2]

I have no reference for this, but I believe that with probability 1 it generates curves of length 1.  I've attached the Mathematica code for a curve in a comment if you want to play with it.


  [1]: https://i.sstatic.net/6fazv.png
  [2]: https://i.sstatic.net/9aSF6.png