Lately I've been generating SVG files of uniform hyperbolic tilings in the Poincaré disc model (PDM). The results are usually very pretty.
To generate snub tilings $sr\{p,q\}$, like $sr\{7,3\}$ above, I need to compute the circumradius of the central polygon (in red above) as seen in the PDM. I do not want to derive this constant numerically, but rather as the root of a polynomial, which is possible because I am not using hyperbolic distances. I have found that, for general $p,q$ with $\frac1p+\frac1q<\frac12$, the circumradius of the central $p$-gon is $\frac{\sqrt x-\sqrt{x-4}}2$ where $x$ is the largest real root of the quartic polynomial $$(a-b)^2x^4\\ +8(a-b)(2a^2-2ab-a+2b)x^3\\ +8(12a^4-18a^3b-11a^3+12a^2b^2+29a^2b+2a^2-24ab^2-12ab+12b^2)x^2\\ +32(8a^5-2a^4b-10a^4+8a^3b^2+8a^3b+3a^3-24a^2b^2-10a^2b+24ab^2+4ab-8b^2)x\\ +16(4a^3+4a^2b-3a^2-8ab+4b)^2$$ and $a=\sin^2\frac\pi p$ and $b=\cos^2\frac\pi q$. A Python implementation can be found here.
My question lies in simplifying the coefficients of this polynomial, themselves polynomials in $a$ and $b$, so to minimise the operation count (and hence time and potential rounding error too) à la this question. The $x^4$, $x^3$ and constant coefficients look simplified enough; I'm most concerned about the $x^2$ and $x$ coefficients. I've been perturbing up to three monomials at once in hopes of getting a neat factorisation of the perturbed polynomial, and this is the best I've got so far: $$8(12a^4+a^3-ab-(3ab+2a-3b)(6a^2-4ab-a+4b))x^2$$ $$32(a^3(4a-3)(2a-1)-2b(a-1)^2(a^2-4ab-2a+4b))x$$
But can I do better than this? Can I reduce the complexity further? Is it possible to leverage intermediate results for the other coefficients in reducing the complexity?