Fourth point of intersection of two conics Five points in general position define a unique conic section. Let $Q_1$ be a conic through points $A,B,C,E_1,F_1$ and likewise $Q_2$ through $A,B,C,E_2,F_2$. Two conics (over an algebraically closed field) generally intersect in four points, so these two will have a fourth point $D$ in common, in addition to the points $A,B,C$ which are common by construction.
What are its coordinates? To be more precise, how can you express the coordinates of that point $D$ as a rational function of the coordinates of the 7 given points $A,B,C,E_1,F_1,E_2,F_2$? And can you certify that the rational functions you found are as simple as possible, in terms of their total degree?
I have an approach to find these coordinates, which I will write up as an answer. But reducing its degree as far as possible makes it really ugly. So I encourage other approaches tackling this differently. If I'm not mistaken, the ideal solution would be of degree 6 in $A,B,C$ and of degree 4 in $E_1,F_1,E_2,F_2$.
I thought of this problem after re-reading a previous answer of mine and being unhappy about the way I translated between Möbius geometry and regular projective geometry there. With $A,B$ taken as the ideal circle points, the problem here can be used to find the second point of intersection for two circles, each defined by three points, and one of them common to both the circles.
 A: Let the conics be $ABCE^+F^+$ and $ABCE^-F^-$, and let the fourth point of intersection be $D$. Write $D$, and the $E$s and $F$s, using rampantly-reciprocated barycentric coordinates
$$D = \left(\;\frac1{a} : \frac1{b} : \frac1{c}\;\right) \qquad E^\pm = \left(\;\frac{1}{a_\circ^\pm} : \frac{1}{b_\circ^\pm} : \frac{1}{c_\circ^\pm} \;\right) \qquad 
F^\pm = \left(\;\frac{1}{a_\star^\pm} : \frac{1}{b_\star^\pm} : \frac1{c_\star^\pm} \;\right)$$ 
(using "$a_\circ^\pm$", etc, to reduce some of the visual clutter otherwise encountered with "$a_E^\pm$") where $(u:v:w)$ represents the point
$$\frac{u A+v B+w C}{u+v+w}$$
Now, a computer algebra system makes pretty quick work of finding the equations for the conics and representations of the points of intersection. I won't bother TeX-ing up any of that here; I'll simply cut to the chase, with a first pass at helpful grouping:
$$\begin{align}
a &= 
\phantom{+} 
  a_\circ^+ a_\circ^- \left(b_\star^+ c_\star^- - c_\star^+ b_\star^- \right)
+ a_\star^+ a_\star^- \left(b_\circ^+ c_\circ^- - c_\circ^+ b_\circ^- \right) \\
&\phantom{=}+ a_\circ^+ a_\star^- \left(b_\circ^- c_\star^+  - c_\circ^- b_\star^+ \right)
+ a_\circ^- a_\star^+ \left( b_\star^- c_\circ^+ - b_\circ^+ c_\star^- \right)
\\[8pt]
b &= 
\phantom{+} 
  b_\circ^+ b_\circ^- \left( c_\star^+ a_\star^- - c_\star^- a_\star^+ \right)
+ b_\star^+ b_\star^- \left( c_\circ^+ a_\circ^- - c_\circ^- a_\circ^+ \right)
\\[4pt]
&\phantom{=}
+ b_\circ^+ b_\star^- \left( c_\circ^- a_\star^+ - c_\star^+ a_\circ^- \right)
+ b_\circ^- b_\star^+ \left( c_\star^- a_\circ^+ - c_\circ^+ a_\star^- \right)
 \\[8pt]
c &= 
\phantom{+} 
  c_\circ^+ c_\circ^- \left( a_\star^+ b_\star^- - a_\star^- b_\star^+ \right)
+ c_\star^+ c_\star^- \left( a_\circ^+ b_\circ^- - a_\circ^- b_\circ^+ \right)
\\ &\phantom{=}
+ c_\circ^+ c_\star^- \left( a_\circ^- b_\star^+ - a_\star^+ b_\circ^- \right)
+ c_\circ^- c_\star^+ \left( a_\star^- b_\circ^+ - a_\circ^+ b_\star^- \right)
\end{align}$$
A: Here is what I have so far, posting this answer together with my question.
Let's start with the property that six points are on a common conic. Using $3\times3$ determinants of homogeneous coordinates, which I'll write using square brackets, that condition is
$$[ACE][BDE][ADF][BCF]-[ADE][BCE][ACF][BDF] = 0$$
I want to describe the unknown point $D$ in terms of the three points $A,B,C$. As long as these are not collinear, you can describe any point in the projective plane as $aA+bB+cC$. But you can also describe almost any point in the plane as
$$D=bcA+acB+abC$$
This works for all points except those where one of the basis points has coefficient zero, i.e. points which are collinear with two of the basis points but identical to none of them.
Now I want to plug this point into the equation of the conic. (Actually it's two equations for the two different conics, but the first steps of the calculations are the same, so I'll not distinguish the $E_i$ and $F_i$ for now.) To make this simpler, I will to split the two occurrences of point $D$ in the equation of the conic, and establish an abbreviated notation for the resulting formula.
$$f(D_1,D_2):=[ACE][BD_1E][AD_2F][BCF]-[AD_1E][BCE][ACF][BD_2F]$$
(If you want you can see this as moving from the quadratic form to a bilinear form, or from the matrix product $D^T\cdot M\cdot D$ to a product $D_1^T\cdot M\cdot D_2$ where $M$ is not even symmetric.) Now plugging the above $D$ into this we get
$$f(D,D)=f(bcA+acB+abC,bcA+acB+abC)=\\
b^2c^2\,f(A,A)+a^2c^2\,f(B,B)+a^2b^2\,f(C,C)+\\
a^2bc\,f(B,C)+a^2bc\,f(C,B)+
ab^2c\,f(A,C)+\\ab^2c\,f(C,A)+
abc^2\,f(A,B)+abc^2\,f(B,A)$$
The first three terms are zero, because $A,B,C$ are on the conic by definition.
All the remaining conics have $abc$ as a common coefficient. Canceling that we get a description which is linear in $a,b,c$.
$$\frac{f(D,D)}{abc}=
a\bigl(f(B,C)+f(C,B)\bigr)+
b\bigl(f(A,C)+f(C,A)\bigr)+
c\bigl(f(A,B)+f(B,A)\bigr)$$
Let'd look at the first of these terms in more detail:
$$f(B,C)=[ACE][BBE][ACF][BCF]-[ABE][BCE][ACF][BCF]$$
Since $[BBE]$ is zero, only one term remains. The same is true for all the other invocations of $f$ here. The sum of two of these can be simplified even further:
$$f(B,C)+f(C,B)=\\
=0-[ABE][BCE][ACF][BCF]+[ACE][BCE][ABF][BCF]-0=\\
=[BCE][BCF]\bigl(-[ABE][ACF]+[ACE][ABF]\bigr)=\\
=-[BCE][BCF][ABC][AEF]$$
This is because $[ABC][AEF]-[ABE][ACF]+[ABF][ACE]=0$ is a Grassmann-Plücker relation: it holds for arbitrary $A,B,C,E,F$. A similar computation can be performed for the other two sums. All of them share $[ABC]$ as a common term, representing a singularity which we can remove by canceling that term. So now we have
\begin{alignat*}{2}
-\frac{f(D,D)}{abc\,[ABC]}=&\,
a[BCE][BCF][AEF]\\
+&\,b[ACE][ACF][BEF]\\
+&\,c[ACE][ABF][CEF]\overset!=0
\end{alignat*}
If we do this for two conics, we get two homogeneous linear equations. Finding their common solution amounts to intersecting two lines in that $a,b,c$ parameter space. Intersecting lines in projective planes translates to cross products. So we get
$$\begin{pmatrix}a\\b\\c\end{pmatrix}=
\begin{pmatrix}
[BCE_1][BCF_1][AE_1F_1]\\
[ACE_1][ACF_1][BE_1F_1]\\
[ABE_1][ABF_1][CE_1F_1]\\
\end{pmatrix}\times
\begin{pmatrix}
[BCE_2][BCF_2][AE_2F_2]\\
[ACE_2][ACF_2][BE_2F_2]\\
[ABE_2][ABF_2][CE_2F_2]\\
\end{pmatrix}$$
Computing $a,b,c$ using this formula and then plugging the result back into $D=bcA+acB+abC$ leads to a deterministic computation for $D$ which is free of divisions, square roots, case distinctions and other annoyances.
The result still has one $[ABC]$ which is removable. But so far I haven't found an elegant way to do this. I can express the bracket ring with all its relations in my computer algebra system Sage, and use that to compute some representation of the resulting polynomial, as a linear combination of at least 4 of the 7 input points. But the corresponding coefficients span several pages and are for this reason rather unsuitable for most applications.
I could confirm that with $A,B,C$ chosen in specific and non-collinear positions, the resulting terms were free of common factors. The choice of positions was without loss of generality, so apart from $[ABC]$ there should be no other removable factors in this result. I also used my CAS to check that $[ABC]^2$ is not a factor either. So unless I made a mistake, the result with $[ABC]$ canceled should indeed be as simple as it gets, in terms of degrees.
I'm still looking for a computation which exhibits the common factor $[ABC]$ in the computation without making the resulting expression unwieldy.
