This answer is an answer for the very specific question:
How to obtain or verify $X(164)$ knowing it is the $X(1)$-point of the excentral triangle?
Explicitly:
How to obtain the barycentric coordinates of $X(164)$, knowing it is the incenter of the circle through the points with (inhomogeneous) barycentric coordinates
$$\begin{aligned}
S&(-a:b:c)\ ,\\
T&(a:-b:c)\ ,\\
U&(a:-b:c)\ ?
\end{aligned}
$$
As a reference, the following collects a lot of information in a beautiful presentation:
Max Schindler, Evan Chen, Barycentric Coordinates for the Impatient
We fix the frame, we will work usually with baricentric normalized coordinates $(x,y,z)$ of a point $P=xA+yB+zC$, considered w.r.t. the triangle $ABC$ with sides $a=BC$, $b=CA$, $c=AB$. The word normalized refers to the relation $x+y+z=1$. While working with points in barycentric coordinates, often expressions get complicated, so it is natural to show the point $(x,y,z)$ only up to the common denominator, which is omitted. In such a case we write $(x:y:z)$, and this point stays for $(\ x/(x+y+z),\ y/(x+y+z),\ z/(x+y+z)\ )$.
The formula for the squared distance $OP^2$ between two points with barycentric normalized coordinates $P(x,y,z)$ and $O(x_0,y_0,z_0)$ is algebraic, explicitly
$$
PO^2
=
-a^2(y-y_0)(z-z_0)
-b^2(x-x_0)(z-z_0)
-c^2(x-x_0)(y-y_0) \ ,
$$
and it factorizes through building the "displacement" (vector) $(x-x_0,y-y_0,z-z_0)$.
In our case, let $S,T,U$ be the ex-centers of $\Delta ABC$. Let us compute first $TU^2$. To have an easy typing, let me please type sage code:
sage: var('a,b,c');
sage: def d2(P, Q):
....: x1, y1, z1 = P
....: x2, y2, z2 = Q
....: x12, y12, z12 = x2-x1, y2-y1, z2-z1
....: return -a^2*y12*z12 -b^2*x12*z12 -c^2*x12*y12
....:
sage: S, T, U = vector([-a,b,c]), vector([a,-b,c]), vector([a,b,-c])
sage: S, T, U = S/sum(S), T/sum(T), U/sum(U)
sage: S, T, U
((a/(a - b - c), -b/(a - b - c), -c/(a - b - c)),
(a/(a - b + c), -b/(a - b + c), c/(a - b + c)),
(a/(a + b - c), b/(a + b - c), -c/(a + b - c)))
sage: d2(T, U).factor()
4*a^2*b*c/((a + b - c)*(a - b + c))
So we can write:
$$
\begin{aligned}
TU^2 &= a^2\cdot \frac{bc}{(p-b)(p-c)}
&&=a^2\cdot\sin^{-2}\frac A2\ ,\text{ and similarly}
\\
US^2 &= b^2\cdot \frac{ac}{(p-a)(p-c)}
&&=b^2\cdot\sin^{-2}\frac B2\ ,
\\
ST^2 &= c^2\cdot \frac{ab}{(p-a)(p-b)}
&&=c^2\cdot\sin^{-2}\frac C2\ .
\end{aligned}
$$
(Above $p=(a+b+c)/2$, since i need $s$ for an other purpose. This is a rare case where i do not use $p$ for a prime number.)
Above it is important that we could write the expressions $TU^2$, $US^2$, $ST^2$ also as squares of "elements" directly connected to the original triangle $\Delta ABC$. (Taking the sine of half an angle brings us slightly out of the world of algebraic expressions in $a,b,c$.) Else i could not continue. The continuation is now clear. Let $s,t,u$ be the sides of $\Delta STU$, so
$$
\begin{aligned}
s & = TU = a\cdot\sin^{-1}\frac A2\ ,
\\
t & = US = b\cdot\sin^{-1}\frac B2\ ,
\\
u & = ST = c\cdot\sin^{-1}\frac C2\ .
\\[3mm]
&\qquad\text{ Then we use the formula for $X(1)$ in $\Delta STU$:}
\\[3mm]
X(1)_{\Delta STU}
&=
\frac 1{s+t+u}(sS+tT+uU)\ ,\text{ and here we plug in }
\\
S &=
\frac{-a}{-a+b+c}A +
\frac{b}{-a+b+c}B +
\frac{c}{-a+b+c}C \ ,\\
\\
T &=
\frac{a}{a-b+c}A +
\frac{-b}{a-b+c}B +
\frac{c}{a-b+c}C \ ,\\
\\
U &=
\frac{a}{a+b-c}A +
\frac{b}{a+b-c}B +
\frac{-c}{a+b-c}C \ ,
\\[3mm]
&\qquad\text{ and we compute of the $A$-coefficient in $X(1)_{\Delta STU}$,}
\\[3mm]
\text{($A$-coefficient)}
&=\frac a{s+t+u}
\left(
\frac{-s}{2(p-a)} +
\frac{t}{2(p-b)} +
\frac{u}{2(p-c)}
\right)
\\
&\sim
a
\left(
-
\frac{a}{(p-a)\sin\frac A2} +
\frac{b}{(p-b)\sin\frac B2} +
\frac{c}{(p-c)\sin\frac B2}
\right)
\\
&=
a
\left(
-
\frac{a}{r\cot\frac A2\sin\frac A2} +
\frac{b}{r\cot\frac B2\sin\frac B2} +
\frac{c}{r\cot\frac C2\sin\frac C2}
\right)
\\
&\sim
a
\left(
-
\frac{\sin A}{\cos\frac A2} +
\frac{\sin B}{\cos\frac B2} +
\frac{\sin C}{\cos\frac C2}
\right)
\\
&\sim
a
\left(
-
\sin\frac A2 +
\sin\frac B2 +
\sin\frac C2
\right)\ .
\end{aligned}
$$
The corresponding coefficients of $A,B,C$ are the needed barycentric coefficients, and they match the one in the ETC.
Above, the symbol $\sim$ denotes equality up to a factor, which is a symmetric polynomial in $a,b,c$.
Conclusion:
The OP reopens implicitly the following door. Given the triangle $ABC$, we associate other triangles $STU$, where $S\cong S(a,b,c)$ is an asymmetric expression, a linear combination in $A,B,C$ with polynomial (or slightly more general) weights in $a,b,c$, maybe symmetric w.r.t. $b\leftrightarrow c$, and $T,U$ are obtained correspondingly using cyclic permutations of $(a,b,c)$ (and implicitly $A,B,C$), so $T\cong S(b,c,a)$, $U\cong S(c,a,b)$.
We compute the squared sides of $STU$ as above, and if the explicit expressions for $TU^2$, $US^2$, $ST^2$ admit a radical, then we can proceed as above to write all centers $X(k)_{STU}$ in terms of $S,T,U$, then using the expressions for $S,T,U$ in terms of $A,B,C$ we obtain weights w.r.t. the initial triangle, and we may want to match them with existing centers.
Application:
Let $STU$ be constructed based on $S=(0,\frac 12,\frac 12)=\frac 12(B+C)$, so $S$ is the mid point of $BC$, and we chose $T,U$ similarly. The sides of $STU$ are $s=a/2$, $t=b/2$, $u=c/2$, so
$$
\begin{aligned}
X(98)_{\Delta STU}
&=
\left(\
\frac 1{t^4+u^4-s^2(t^2+u^2)}\ :\
\dots
\ \right)
\\
&=
\left(\
\frac 1{b^4+c^4-a^2(b^2+c^2)}\ :\
\dots
\ \right)
\\
&\sim \frac 1{b^4+c^4-a^2(b^2+c^2)}S+\dots
\\
&\sim \frac 1{b^4+c^4-a^2(b^2+c^2)}(B+C)+\dots
\\
&\sim
\left(
\frac 1{a^4+b^4-c^2(a^2+b^2)}+
\frac 1{a^4+c^4-b^2(a^2+c^2)}
\right)A+\dots
\\
&\sim
(b^4+c^4-a^2(b^2+c^2))
\cdot\Big(\
(a^4+b^4-c^2(a^2+b^2))+
(a^4+c^4-b^2(a^2+c^2))
\ \Big)A+\dots
\\
&\qquad\text{ matching again the ETC, and giving}
\\
&=X(114)_{\Delta ABC}\ .
\end{aligned}
$$
Since the circumference is the place of the points distant $R > 0$ from a point $\left(x_c, \, y_c\right)$, its cartesian equation is:
$$
\left(x - x_c\right)^2 + \left(y - y_c\right)^2 = R^2
$$
whose rational parameterization is $\left(x, \, y\right) = \left(x_c + R\,\frac{1-z^2}{1+z^2}, \, y_c + R\,\frac{2\,z}{1+z^2}\right)$, with $z \in \mathbb{R}$.
At this point, an ellipse assigned in canonical form:
$$\frac{\left(x - x_c'\right)^2}{a^2} + \frac{\left(y - y_c'\right)^2}{b^2} = 1$$
with $a, \, b > 0$, substituting for $x, \, y$ the respective parametric equations of the circumference is obtained:
$$
\frac{\left(x_c + R\,\frac{1-z^2}{1+z^2} - x_c'\right)^2}{a^2} +
\frac{\left(y_c + R\,\frac{2\,z}{1+z^2} - y_c'\right)^2}{b^2}
= 1
$$
ie:
$$
\small
\left(x_c - x_c' + R\,\frac{1-z^2}{1+z^2}\right)^2\,b^2\left(1+z^2\right)^2 +
\left(y_c - y_c' + R\,\frac{2\,z}{1+z^2}\right)^2\,a^2\left(1+z^2\right)^2
= a^2\,b^2\left(1+z^2\right)^2
$$
from which:
$$
\small
b^2\left(x_c - x_c'\right)^2\left(1 + z^2\right)^2 +
2\,b^2\,R\left(x_c - x_c'\right)\left(1 - z^4\right) +
b^2\,R^2\left(1 - z^2\right)^2 + \\
\small
a^2\left(y_c - y_c'\right)^2\left(1 + z^2\right)^2 +
2\,a^2\,R\left(y_c - y_c'\right)\left(2\,z\right)\left(1 + z^2\right) +
a^2\,R^2\left(2\,z\right)^2
= a^2\,b^2\left(1+z^2\right)^2
$$
which simplified is equivalent to the equation:
$$
a_4\,z^4 + a_3\,z^3 + a_2\,z^2 + a_1\,z + a_0 = 0
$$
with:
$$
\begin{aligned}
& a_4 = a^2\left(\left(y_c - y_c'\right)^2 - b^2\right) + b^2\left(\left(x_c - x_c'\right) - R\right)^2 \\
& a_3 = 4\,a^2\,R\left(y_c - y_c'\right) \\
& a_2 = 2\left(a^2\left(\left(y_c - y_c'\right)^2 - b^2 + 2\,R^2\right) + b^2\left(\left(x_c - x_c'\right)^2 - R^2\right)\right) \\
& a_1 = 4\,a^2\,R\left(y_c - y_c'\right) \\
& a_0 = a^2\left(\left(y_c - y_c'\right)^2 - b^2\right) + b^2\left(\left(x_c - x_c'\right) + R\right)^2
\end{aligned}
$$
All that remains is to solve this equation and replace the real solutions in the parametric equations of the circumference to obtain the coordinates of the desired points of intersection.
Furthermore, in the event of:
$$
\Delta := \left(\left(x_c - x_c'\right) - (R - a)\right)\left((R + a) - \left(x_c - x_c'\right)\right) \ge 0
$$
and:
$$
a\left(y_c - y_c'\right) - b\,\sqrt{\Delta} = 0
\; \; \; \vee \; \; \;
a\left(y_c - y_c'\right) + b\,\sqrt{\Delta} = 0
$$
then the circumference and the ellipse intersect in the singular point $\left(x_c - R, \; y_c\right)$.
This algorithm can be implemented relatively easily in Microsoft Excel as follows:
referring, for example, to what is reported in the following excellent books:
- Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables - Abramowitz, Stegun - USA
- Handbook of Mathematical Functions - Frank W. J. Olver - NIST
Best Answer
In general if you are given an explicit formula for a polynomial $f(x,y):=\sum_{(i,j)} a_{i,j}x^iy^j \in k[x,y]$ where $k$ is a subfield of the field $\mathbb{C}$ of complex numbers, it is difficult to determine if $f$ is an irreducible polynomial in $k[x,y]$. If the ideal $I:=(f)\subseteq k[x,y]$ is non-prime, it is difficult to calculate the irreducible components $f=f_1\cdots f_k$ of $f$. I dont think there are any "general methods". How do you check that the polynomial is "absolutely irreducible" on a computer?
Note: A computer has "finite memory", hence when you calculate on a computer you calculate in a finite field with large characteristic. So if you have an algorithm on your computer that claims to "prove a polynomial is irreducible", this algorithm proves it is irreducible in a polynomial ring over a finite field.
Example: Let $A:=\mathbb{Z}[x_1,..,x_n]$ be the polynomial ring in $n$ variables over the integers and assume $p>0$ is the largest prime expressible on your computer. Let $k:=\mathbb{Z}/p\mathbb{Z}$.
Assume $F(x_1,..,x_n):=f_1(x_i)\cdots f_k(x_i)+pf(x_i)\in A$ is an irreducible polynomial. Your computer is working in the polynomial ring $B:=k[x_1,..,x_n]$ and in this ring (when reducing $F$ modulo the prime $p$) you get the polynomial
$$ \overline{F}=\overline{f_1}(x_i)\cdots \overline{f_k}(x_j),$$
and the polynomial $\overline{F}$ is not irreducible in $B$. Hence the original polynomial $F$ is irreducible in $A$ but your computer does not detect this since it is working with the polynomial ring $B$.
Comment: "This curve appears to be decomposable into four parts, one of which is the one I want. The problem is that the polynomial is absolutely irreducible, something I have checked with my implementation of Gao's factorisation algorithm (one of whose steps I inquired about in this question). I want to eliminate the unwanted branches of this curve, but factoring does not seem to be the way."
In general: A computer cannot detect if a polynomial is irreducble in $\mathbb{Z}[x_i]$ or $\mathbb{Q}[x_i]$ because it has "finite memory" and is working with polynomials in a polynomial ring over a finite field.
Example: Let $p>0$ be a prime number and consider the polynomial $$f(x):=x^2+p\in A:=\mathbb{Q}[x].$$
Let $p$ be the largest prime expressible on your computer. You can yourself verify that $f(x)$ is irreducible in $A$ but your computer believes there is an equality $f(x)=x^2$ which is a reducible polynomial, since your computer calculates in the field $\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z}$ and the polynomial ring $\mathbb{F}_p[x]$.
Here is a post on the problem of studying the number of solutions to polynomial equations and computer algebra:
When does a system of polynomial equations have infinitely many solutions?