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The question is related to this question:

In this handout theorem 2.3(Homographies that exist) item two claims

For some circle ω and interior points P, Q, we can send ω to itself and send P to Q. (Note that Q is usually taken to be the center of ω.).

I attempt to prove this. It suffices to prove that every interior point can be mapped to center (so the composition of a transform with an inverse transform will send the first interior point to the second interior point).

Let A be the center of ω. Take an interior point B of ω.

Let CD be the diameter through B.

Let EF be the chord perpendicular to CD at A.

Let GH be the chord perpendicular to CD at B.

By theorem 2.3 item one, there exists a projective transformation $T$ on $\Bbb{RP}^2$ that maps C,D,G,H to C,D,E,F respectively.

Then $T$ sends $B$ to $A$.

It remains to prove $T$ sends the circle ω to itself.

But I don't know how to prove $T$ sends the circle ω to itself.

hbghlyj
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1 Answers1

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The unit circle has paramatrization $\left(\frac{1-t^2}{t^2+1},\frac{2 t}{t^2+1}\right)$

In this paramatrization, $C,D,E,F,G,H$ correspond to $1,-1,\infty,0,\frac1x,x$ respectively.

The projective transformation $t\mapsto\frac{x-t}{xt-1}$ maps $1,-1,\frac1x,x$ to $1,-1,\infty,0$.

So there is a projective transformation $T$ on the unit circle that maps $C,D,G,H$ to $C,D,E,F$ respectively.

Another way:

The projective transformation $t\mapsto\frac{t-x}{xt-1}$ maps $1,-1,\frac1x,x$ to $-1,1,\infty,0$.

So there is a projective transformation $T$ on the unit circle that maps $C,D,G,H$ to $D,C,E,F$ respectively.

Moreover, $\left(\begin{smallmatrix}1&-x\\x&-1\end{smallmatrix}\right)$ has trace $0$, so $T$ is an involution. The involution center $S$ is the intersection of lines $FG,EH$.

hbghlyj
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