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Jun 18, 2022 at 10:05 history edited Fedor Petrov CC BY-SA 4.0
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Jun 18, 2022 at 3:04 comment added paul garrett A surprising and elegant solution! But, as in the other comments, probably human-computation-intractable! Interesting... :)
Jun 18, 2022 at 1:00 comment added Thomas Blok @Gro-Tsen: The symmetry constraint forces the off diagonal terms to be the other intersection points of the circles. I would guess the degree 4 factors correspond to when the triple intersection is one of those points. The final factor of degree 6 and number of terms 720 makes me suspect there is some 6x6 matrix determinant in the background. If you would be willing to do one further simplification... Could you plug in the matrix of coordinates that I have given in my question edit? Perhaps it factorizes further. I'm hoping to arrive at a solvable diophantine equation in a, b, x, y.
Jun 17, 2022 at 22:03 comment added Gro-Tsen @ThomasBlok: Yes, with the additional symmetry constraint, the polynomial factorizes into $4$ factors (I imagine only the largest is of any interest). The Sage code and results are here (you're probably only interested in the last four lines, or even the very last).
Jun 17, 2022 at 21:16 comment added Thomas Blok Is there a way to represent this with a larger determinant in terms of the coordinates? @Gro-Tsen, if we were to use the constraint that the 3x3 matrix of coordinates is symmetric, does the formula simplify/factorize?
Jun 16, 2022 at 8:24 comment added Gro-Tsen If you write this out fully, this gives an impossibly complicated expression in the $3\times 3\times 2$ point coordinates, however. (Or at least, a variation I tried, writing the circle equations in the form $x^2 + y^2 + ux + vy + w = 0$, solving for $u,v,w$ and using elimination to find a condition for three circles to meet, gave something impossibly complicated, see here.) We are left to wonder if this expression can be factored and/or simplified.
Jun 15, 2022 at 13:46 history answered Fedor Petrov CC BY-SA 4.0