Problem. Show that for two circles $\omega_1,\omega_2$, the locus of points which are of a constant power ratio to both circles is also a circle (or line) coaxal to the both circles.

Solution 1. First we note that the locus is a circle. For some $k$, the locus we want is all points $X$ satisfying $$p(X,\omega_1)^2 - k^2\cdot p(X,\omega_2)^2 = 0$$ and this is a quadratic polynomial in $(x_X,y_X)$.

Solution 2. Use $(a,b,ir)$ to lift both circles to points $W_1,W_2$. In particular, for a real point $X=(a,b,0)$, the power $p(X,\omega_i)$ is precisely $|XW|^2$ (if you define the distance $|(x_1,x_2,ix_3)| = \sqrt{x_1^2 + x_2^2 - x_3^2}$).

If you now believe that geometry still works in this weird version of $\mathbb R^3$, then you can draw the Apollonius circle (cylinder?), so there are points $H_1,H_2$ such that $$\frac{H_1W_1}{H_1W_2} = \frac{H_2W_2}{H_2W_2}$$ so $(W_1,W_2; H_1,H_2)$ are harmonic, and all the points we want are on the circle with diameter $H_1H_2$, but this also becomes a circle when we just restrict to $\mathbb R^2$.

As for coaxal (hmm), maybe it’s preserved? idk yet. Let’s think more about this.

Discussion.

• Can we prove Casey’s theorem using this?

  Yes! The distance between the $(a,b,ir)$ points is precisely the length of the common external tangent. So it remains to check that the four points are concyclic. One approach is to take the exsimilcenter (so the exsimilicenter of $X,Y$ is simply the intersection of $XY$ and $\{r=0\}$), then by checking power.

• 2+1 spacetime anyone? How do we know that “geomtery” still works?

  We can ask a similar question: why does geometric facts (e.g. power of a point etc.) work in $\mathbb R^3$? Because a single slice is like $\mathbb R^2$.

Remark. Most useful situation is: if you have $\omega$ tangent to $\Gamma$ at $T$, then for any point $X\in \Gamma$, $\frac{p(X,\omega)}{XT^2}$ is a constant.

To illustrate the power of this, here’s a “difficult” mixtilinear fact proven using this easily:

Problem. (CTST?) Let $ABCD$ be a cyclic quadrilateral and let $\omega$ be a circle tangent to $AC$ at $E$, $BD$ at $F$ and arc $CD$ at $T$. Let $M$ be the midpoint of arc $CD$. Show that $CD,EF,TM$ are concurrent.

Solution. Note that $\frac{CE}{CT} = \frac{DF}{DT}$, then we use angle bisector theorem + Menelaus.

Additionally, reminder that we can prove the parallel tangent lemma using Casey’s (or more low tech: Casey with two point-circles is just Ptolemy + the power ratio observation).

Problem. (Parallel tangent lemma) Given $\omega_1,\omega_2$ (internally) tangent to $\Gamma$, let $\ell$ be an external tangent to $\omega_1$, and let $\ell_1,\ell_2$ be the internal tangents to $\omega_1,\omega_2$, intersecting the arc cut out by $\ell$ (not containing the tangency points) at $A_1,A_2$. Then $\ell // A_1A_2$.

Sketch. Let $M$ be the midpoint of that arc. Ptolemy on $A_\bullet MT_1T_2$ to get the length of $A_\bullet M$ as a symmetric quantity.

Answer to: How do we know that “geometry” still works in weird spaces like $\mathbb R\times \mathbb R \times i\mathbb R$?

Firstly, we do a reduction to $\mathbb C\times \mathbb C$, since we only ever care about a 2-dimensional subspace (like even in $\mathbb R^3$, we will restrict to a plane and do planar geometry there).

So the most basic case of our concerns should be explored on $\mathbb R \times i\mathbb R$, say. Now, distance is described by $d(x,y) = x^2-y^2$, so our “circles” become rectangular hyperbolas.

(If you imagined each circle as secretly being the cross-section of a cone, this is the same picture but tilted 90-degrees into the $z$-axis.)

Why does geometry still work then?

First let’s think about what we mean by geometry. Ptolemy’s theorem says that for any 4 points on a circle, some polynomial in terms of the coordinates must equal to 0. In equations, something like: $$ \begin{aligned} (x_1-a)^2 + (y_1-b)^2 = R^2\\ (x_2-a)^2 + (y_2-b)^2 = R^2\\ (x_3-a)^2 + (y_3-b)^2 = R^2\\ (x_4-a)^2 + (y_4-b)^2 = R^2 \end{aligned}\Rightarrow P(x_1,y_1,...,x_4,y_4) = 0 $$ where $x_i,y_i\in \mathbb R$, and perhaps we would like to extend this to $x_i,y_i\in \mathbb C$.

Is that always true? No, consider for instance if $R=0$, then the LHS condition (over $\mathbb R$) is that $x_i=y_i=0$ implies $P(x)=0$ (which is extremely weak), and clearly that doesn’t mean that the $\mathbb C$-version should be true because there’s a lot more points there.

A relevant reference: On the Relation between Real Euclidean and Complex Projective Geometry

The eventual answer is: as long as the polynomial conditions aren’t degenerate, this is true! Rigorously, first we define:

A locus is the solution set to one or more polynomials (simultaneously).

Thm. Suppose an “irreducible” locus $L$ has the same dimension over $\mathbb R$ as over $\mathbb C$. Then any locus containing $L(\mathbb R)$ must also contain $L(\mathbb C)$.

There’s a bunch of things not properly defined, but essentially this nukes the question of whether the same “geometric” facts hold in $\mathbb C^2$-planes.

How are things like these proven? The answer would be actual algebraic geometry.

Comments.

• In $\mathbb R\times i\mathbb R$ the circle points actually show up as points at infinity!
• I wonder how angles work in $\mathbb C^2$…