Context: more stuff left on the cutting floor for my Algebra (intersects) Geometry session handout.
Because Brianchon’s theorem can be proved using radical axes, taking the duals gives the following:
Pascal’s Theorem. Let $ABCDEF$ be a hexagon inscribed in a circle. Show that $AB\cap DE, BC\cap EF, CD\cap FA$ are collinear.
Solution. The idea is to construct three circles $\Omega_1;\Omega_2;\Omega_3$ passing through $A,D;B,E;C,F$ respectively such that the exsimilcenter of $\Omega_1,\Omega_2$ is $AB\cap DE$ (and analogous relations for the other two pairs).
We claim that it suffices to take the $\Omega_i$’s to be orthogonal to $(ABCDEF)$. So it reduces to the following observation:
Observation: Given circles $\omega_1, \omega_2$, any circle orthogonal to both $\omega_1,\omega_2$ is fixed by the inversion swapping $\omega_1$ and $\omega_2$. Proof: Preservation of angles, so the image of $\gamma$ is still orthogonal to both circles. Then there should be a unique circle perspective from the center of inversion that is orthogonal to both circles. (Or if you don’t buy this, invert the two circles onto intersecting great circles.)
This implies that the inversion at the exsimilcenter swaps $A\leftrightarrow B$ and $D\leftrightarrow E$, so we’re done.
Remark. We don’t need orthogonality exactly: just that these circles form the same angle $\theta$ with $(ABCDEF)$. So there’s actually one variable real parameter, just like the Brianchon proof (!).
Additionally, I don’t know if there are sign issues we have to deal with.
Also some other interesting stuff: