@Noam: Drat, yes. It was rather presumptuous of me to assume a simple solution could be described, after the heroic efforts of people like Blue who have already replied (in the mathoverflow thread in his case). I'll say no more unless I can devise a formal solution!

@mathlove: Flip, yes you're correct, and I don't see any way to rescue it! In that case the solution must be to draw your line through the intersection of the two starting lines at such an angle to ensure (given the chord lengths known initially) that the chord length of an intersection, if any, is unique to the side of the "cross" formed by the pair of initial lines, whatever the circle diameter. If there is no intersection, then try the same trick with the other pair of angles. (It's obvious one of these lines must clip the circle.)

Many thanks for your reply, David, which I've marked as the definive one, and thanks to everyone else who replied. I should have remembered that "real gap" observation, as this is a text book example!

Not sure if it is relevant to your question (hence comment rather than reply), but the simultaneous pair $x + y + z + x y z = 0, x^2 + y^2 = z^2 + t^2$ is birationally equivalent to the set $1 + X^2 = U^2, 1 + Y^2 = V^2, 1 + X^2 + Y^2 = W^2$, and to many other "interesting looking" sets of one or more equations defining a surface, including $1 + X^2 Y^2 = (X^2 + Y^2) Z^2$

That argument works, but with mod $p$, where $p$ is the smallest prime dividing $n$

Can someone briefly explain why this was marked down? (N.B. I'm not disputing the reason one way or another, just curious)