There seems to be an obvious geometric approach. Let T be the 3-torus, and take the smooth function F on it to $\mathbb{R}^3$ like this: use three angles on three given circles as the parameters on T, and from the points P, Q, R on the respective circles construct F as the squares of the Euclidean distances from P to Q, Q to R, R to P. So we are interested in the cases where F maps a point of T to the lattice point (1, 1, 1). The inverse image of the lattice point is a closed subset of T. To make it finite, we need by compactness of T only to understand the derivative of F: where it is invertible the inverse function theorem will work for us. So it seems to come down to computing the derivative of F, in explicit terms of the centres and radii of the circles. (The margin here is too small for so much notation.)

**Edit**: I now understand the problem a bit better, having manipulated the Jacobian of F. It appears to vanish under the following (sufficient) condition. Write x(1), x(2) and x(3) for the centres of the circles, and v(1) etc. for the corresponding "velocity vectors", i.e. the tangent vectors at a given point of a circle of length given by the radius, which are what one finds as the derivative of the position of a rotating point. Key quantities are the scalar products (x(1) - x(2)).(v(1) - v(2)), and so on. Where all three of these scalar products vanish, the derivative of F is not invertible. This can be seen to happen in particular configurations where the centres are at the vertices of an equilateral triangle, and the circles have equal radius. This condition is not clearly necessary, however.