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.

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:
$$\begin{align}
a &= 
\phantom{+} 
  a_\circ^+ a_\circ^- \left(b_\star^+ c_\star^- - c_\star^+ b_\star^- \right)
+ a_\star^+ a_\star^- \left(b_\circ^+ c_\circ^- - c_\circ^+ b_\circ^- \right) \\
&\phantom{=}+ a_\circ^+ a_\star^- \left(b_\circ^- c_\star^+  - c_\circ^- b_\star^+ \right)
+ a_\circ^- a_\star^+ \left( b_\star^- c_\circ^+ - b_\circ^+ c_\star^- \right)
\\[8pt]
b &= 
\phantom{+} 
  b_\circ^+ b_\circ^- \left( c_\star^+ a_\star^- - c_\star^- a_\star^+ \right)
+ b_\star^+ b_\star^- \left( c_\circ^+ a_\circ^- - c_\circ^- a_\circ^+ \right)
\\[4pt]
&\phantom{=}
+ b_\circ^+ b_\star^- \left( c_\circ^- a_\star^+ - c_\star^+ a_\circ^- \right)
+ b_\circ^- b_\star^+ \left( c_\star^- a_\circ^+ - c_\circ^+ a_\star^- \right)
 \\[8pt]
c &= 
\phantom{+} 
  c_\circ^+ c_\circ^- \left( a_\star^+ b_\star^- - a_\star^- b_\star^+ \right)
+ c_\star^+ c_\star^- \left( a_\circ^+ b_\circ^- - a_\circ^- b_\circ^+ \right)
\\ &\phantom{=}
+ c_\circ^+ c_\star^- \left( a_\circ^- b_\star^+ - a_\star^+ b_\circ^- \right)
+ c_\circ^- c_\star^+ \left( a_\star^- b_\circ^+ - a_\circ^+ b_\star^- \right)
\end{align}$$
