How can I obtain a polynomial formula for the product (i.e. union) of a triangle's incircle and excircles, avoiding the use of radicals?
Assume three lines in general position are given by three equations
$$a_ix+b_iy+c_i=0\quad\text{for }i\in\{1,2,3\}$$
A circle with center $(p,q)$ and radius $r$ may be given by the equation
$$(x-p)^2+(y-q)^2-r^2=0$$
which can also be written as
$$(x,y,1)\begin{pmatrix}1&0&-p\\0&1&-q\\-p&-q&p^2+q^2-r^2\end{pmatrix} \begin{pmatrix}x\\y\\1\end{pmatrix}=0$$
Using the adjugate matrix to switch from primal to dual conic, we can say a line is tangent to that circle if it satisfies
$$(a,b,c)\begin{pmatrix} p^2-r^2 & pq & p \\ pq & q^2-r^2 & q \\ p & q & 1 \end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}=0$$
So by plugging in the three lines from the start, we get three non-linear equations. In general we get 4 distinct solutions for $p,q,r^2$. (We actually get 8 solutions if we solve for $r$ instead of $r^2$, but these are only $r=\pm\sqrt{r^2}$ and by convention we'd pick the positive radius. All the formulas only use $r^2$ not $r$ so it makes sense to treat that as the variable.) These 4 solutions correspond to the incircle and the three excircles of the triangle formed by the lines.
The underlying quartic equation behind this would introduce plenty of radicals, and rules on how to match the solutions, which complicates subsequent work. It would be much better if we could avoid taking roots here. And I think that by dealing with all four circles together, that should be possible. Specifically I'm looking for the product of the four circle equations, which corresponds to an algebraic curve of degree 8 that represents the union of the four circles. I conjecture that the product of the four circles, i.e. the formula
$$0=\prod_{i=1}^4 (x-p_i)^2+(y-q_i)^2-r_i^2$$
can be stated as a polynomial of combined degree 8 in $x,y$ where the coefficients themselves can be given as polynomials in the coordinates of the lines, namely $a_i,b_i,c_i$. If the coordinates of the lines are rational, then the coefficients in the product of circles will be rational, too.
I have checked the last part of this conjecture, the rationality of coefficients, using one specific triangle, chosen fairly arbitrarily (using three rational points on the unit circle):
\begin{align*} a_1 &= 23 & b_1 &= 41 & c_1 &= -47 \\ a_2 &= 4 & b_2 &= -7 & c_2 &= 4 \\ a_3 &= 3 & b_3 &= -5 & c_3 &= 3 \end{align*}
For this I got four circles characterized by
\begin{align*} 244205p^4 - 366418p^2 + 243984p - 45648 &= 0 \\ 244205q^4 - 1098812q^2 + 1268540q - 411845 &= 0 \\ 59636082025r^8 - 598843408280r^6 + 529183150574r^4 - 356490680r^2 + 60025 &= 0 \end{align*}
Matching the correct roots of these polynomials to get consistent solutions:
\begin{align*} p_1 &\approx \phantom+0.47089 & q_1 &\approx \phantom+0.86139 & r_1^2 &\approx 0.00032872 \\ p_2 &\approx \phantom+0.50761 & q_2 &\approx \phantom+0.88289 & r_2^2 &\approx 0.00034533 \\ p_3 &\approx -1.49989 & q_3 &\approx \phantom+0.85359 & r_3^2 &\approx 0.97839603 \\ p_4 &\approx \phantom+0.52138 & q_4 &\approx -2.59788 & r_4^2 &\approx 9.06255888 \end{align*}
The product of these four circles, scaled to avoid divisions, is then the following:
\begin{align*} 59636082025&\,x^8 \\ + 238544328100&\,x^6y^2 \\ + 357816492150&\,x^4y^4 \\ + 238544328100&\,x^2y^6 \\ + 59636082025&\,y^8 \\ - 241134854740&\,x^6 \\ + 424846368960&\,x^5y \\ - 1438821671300&\,x^4y^2 \\ + 849692737920&\,x^3y^3 \\ - 2154238778380&\,x^2y^4 \\ + 424846368960&\,xy^5 \\ - 956551961820&\,y^6 \\ + 108802118880&\,x^5 \\ + 265528980600&\,x^4y \\ - 1771920221760&\,x^3y^2 \\ + 3788213456560&\,x^2y^3 \\ - 1880722340640&\,xy^4 \\ + 3522684475960&\,y^5 \\ - 12729722876&\,x^4 \\ + 1691695948800&\,x^3y \\ - 2241047404232&\,x^2y^2 \\ + 2683644936960&\,xy^3 \\ - 6476277331836&\,y^4 \\ - 484880944704&\,x^3 \\ + 468260157040&\,x^2y \\ - 1625087765184&\,xy^2 \\ + 7083496190160&\,y^3 \\ + 48256725504&\,x^2 \\ + 293698179840&\,xy \\ - 4644695250832&\,y^2 \\ - 13129818624&\,x \\ + 1675829775360&\,y \\ - 243236814336&\quad=0 \end{align*}
This polynomial will factor into four circles over $\mathbb R$ or $\mathbb A$ but is irreducible over $\mathbb Q$.
How can I get these coefficients without going through the detour of the four distinct circles and their irrational parameters? How can I do this in situations where the coordinates of the lines themselves might contain unknowns?
Background for my question is Probability that the centroid of a triangle is inside its incircle. For an exact solution there, radicals would make one's life really hard. But at the same time, since the centroid will never lie within an excircle, considering the sign of the product of circles should work just as well, and when combined with a rational parametrization of the circle might lend itself to some nice algebraic approach for that question. That's what got me thinking, but at the moment I'm actually more intrigued by this question here for its own merit. I feel like I'm missing some very useful tool in my arsenal, but don't know how to learn more.
I'm including the Galois theory tag as I have the rough understanding that Galois theory deals with the relationship between the different roots of a polynomial. So if I want to understand how the different solutions interact when I multiply the circles, I assume that topic might have contributions. But so far my knowledge of Galois theory is pretty much exhausted by getting my computer algebra system to compute the Galois group of some polynomial, and then using that to decide whether a number is constructible or not.
Update: For a moment I thought that the tool I had missed might be Vieta's formulas. But some of my coefficients will be combinations of roots of different polynomials. So I can't predict all my coefficients by just looking at the defining polynomial for one of my circle parameters, and when I look at two then the problem of how to match the roots returns. It seems to me that just the two polynomials won't have enough information on how to do that.
To expand on this idea: If we define
$$d_i:=-2p_i\qquad e_i:=-2q_i\qquad f_i:=p_i^2+q_i^2-r_i^2$$
then the product of circles is
$$0 = \prod_{i=1}^4 (x+y)^2 + d_ix + e_iy + f_i$$
By treating $(x^2+y^2)$ as a single variable we can achieve a form where each monomial has a more direct correspondence with the $3\times4$ coefficients from the individual circles:
\begin{align*} 0 = \Bigl(59636082025 &\, (x^2+y^2)^4 \\ -\,357924430760 &\, (x^2+y^2)^2x^2 \\ +\,476656901760 &\, (x^2+y^2)x^3 \\ -\,178359517440 &\, x^4 \\ +\,424846368960 &\, (x^2+y^2)^2xy \\ -\,778885009760 &\, (x^2+y^2)x^2y \\ +\,353718243840 &\, x^3y \\ -\,1073341537840 &\, (x^2+y^2)^2y^2 \\ -\,1512867557760 &\, (x^2+y^2)xy^2 \\ +\,2460406401440 &\, x^2y^2 \\ +\,2478270485600 &\, (x^2+y^2)y^3 \\ +\,1345667232000 &\, xy^3 \\ -\,1609193731600 &\, y^4 \\ +\,116789576020 &\, (x^2+y^2)^3 \\ -\,367854782880 &\, (x^2+y^2)^2x \\ +\,354562216320 &\, (x^2+y^2)x^2 \\ -\,103527290880 &\, x^3 \\ +\,1044413990360 &\, (x^2+y^2)^2y \\ +\,1337977704960 &\, (x^2+y^2)xy \\ -\,2253031422720 &\, x^2y \\ -\,4678151178480 &\, (x^2+y^2)y^2 \\ -\,1243734111360 &\, xy^2 \\ +\,4362204610400 &\, y^3 \\ -\,188932421756 &\, (x^2+y^2)^2 \\ -\,381353653824 &\, (x^2+y^2)x \\ +\,537823817856 &\, x^2 \\ +\,2721291579760 &\, (x^2+y^2)y \\ +\,293698179840 &\, xy \\ -\,4155128158480 &\, y^2 \\ -\,489567092352 &\, (x^2+y^2) \\ -\,13129818624 &\, x \\ +\,1675829775360 &\, y \\ -\,243236814336 & \Bigr)/59636082025 \end{align*}
We can get the parameters of the individual circles characterized by
\begin{align*} 244205d^4 - 1465672d^2 - 1951872d - 730368 &= 0 \\ 244205e^4 - 4395248e^2 - 10148320e - 6589520 &= 0 \\ 59636082025f^4 - 116789576020f^3 - 188932421756f^2 + 489567092352f - 243236814336 &= 0 \end{align*}
Now for example the coefficient of $x^4$ is
$$d_1d_2d_3d_4=-\frac{730368}{244205}=-\frac{178359517440}{59636082025}$$
This is a straight-forward application of Vieta, dividing the constant last coefficient by the leading one. Similarly the coefficient for $(x^2+y^2)^3$ is
$$f_1+f_2+f_3+f_4=\frac{2164}{1105}=\frac{116789576020}{59636082025}$$
again following straight from the polynomial for $f$, dividing the second coefficient by the leading one and flipping the sign. But other coefficients require miyed combinations of letters $d,e,f$. For example the one for $(x^2+y^2)^2xy$ is
$$\sum_{i=1}^4\sum_{\substack{j=1\\j\neq i}}^4d_ie_j=\frac{7872}{1105}=\frac{424846368960}{59636082025}$$
I guess I might be able to somehow use
$$\sum_{i=1}^4\sum_{\substack{j=1\\j\neq i}}^4d_ie_j= \left(\sum_{i=1}^4d_i\right)\left(\sum_{i=1}^4e_i\right)-\left(\sum_{i=1}^4d_ie_i\right)$$
if I were to also compute the defining quartic polynomial for the product $d_ie_i$, but if I get a polynomial for the product by combining the polynomials for $d$ and $e$ then I get a degree $4\times 4=16$ which has all the combinations. In my example I find
$$59636082025(de)^4 + 424846368960(de)^3 - 318329227424(de)^2 - 3822912737280(de) + 4812774543360 = 0$$
is actually quartic, but I don't know how to find this polynomial without going through the irrational individual circles first. For some of the other coefficients from the big equation, the inclusion-exclusion formulas I need to get all the combinations of letters with different indices would be even more complicated.
