12.3: Auxiliary statements
 Page ID
 23656
One may compare the conformal model with a telescope — it makes it possible to see the hplane from the Euclidean plane. Continuing this analogy further, we may say that the following lemma will be used to aim the telescope at any particular point in the hplane.
Lemma \(\PageIndex{1}\)
Consider an hplane with a unit circle as the absolute. Let \(O\) be the center of the absolute and \(P\) be another hpoint. Suppose that \(P'\) denotes the inverse of \(P\) in the absolute.
Then the circle \(\Gamma\) with the center \(P'\) and radius \(\dfrac{\sqrt{1OP^2}}{OP}\) is perpendicular to the absolute. Moreover, \(O\) is the inverse of \(P\) in \(\Gamma\).
 Proof

Follows from Exercise 10.5.2.
Assume \(\Gamma\) is a circline that is perpendicular to the absolute. Consider the inversion \(X \mapsto X'\) in \(\Gamma\), or if \(\Gamma\) is a line, set \(X \mapsto X'\) to be the reflection across \(\Gamma\).
The following observation says that the map \(X \mapsto X'\) respects all the notions introduced in the previous section. Together with the lemma above, it implies that in any problem that is formulated entirely in hterms we can assume that a given hpoint lies in the center of the absolute.
Theorem \(\PageIndex{1}\)
The map \(X \mapsto X'\) described above is a bijection from the hplane to itself. Moreover, for any hpoints \(P\), \(Q\), \(R\) such that \(P\ne Q\) and \(Q\ne R\), the following conditions hold:

The hline \((PQ)_h\), hhalfline \([PQ)_h\), and hsegment \([PQ]_h\) are transformed into \((P'Q')_h\), \([P'Q')_h\), and \([P'Q']_h\) respectively.

\(\delta(P',Q')=\delta(P,Q)\) and \(P'Q'_h=PQ_h\).

\(\measuredangle_h P'Q'R' \equiv \measuredangle_h PQR\).
It is instructive to compare this observation with Proposition [prop:reflection].
 Proof

According to Theorem 10.5.1, the map sends the absolute to itself. Note that the points on \(\Gamma\) do not move, it follows that points inside of the absolute remain inside after the mapping. Whence the \(X \mapsto X'\) is a bijection from the hplane to itself.
Part (a) follows from Theorem 10.3.1 and Theorem 10.6.1.
Part (b) follows from Theorem 10.2.1.
Part (c) follows from Theorem 10.6.1.
Lemma \(\PageIndex{2}\)
Assume that the absolute is a unit circle centered at \(O\). Given an hpoint \(P\), set \(x=OP\) and \(y=OP_h\). Then
\(y = \ln \dfrac{1 + x}{1  x}\) and \(x = \dfrac{e^y  1}{e^y + 1}.\)
Observe that according to lemma, \(OP_h \to \infty\) as \(OP \to 1\). That is if \(P\) approaches absolute in Euclidean sense, it escapes to infinity in the hsense.
 Proof

Note that the hline \((OP)_h\) forms a diameter of the absolute. If \(A\) and \(B\) are the ideal points as in the definition of hdistance, then
\(\begin{array} {l} {OA = OB = 1,} \\ {PA = 1 + x,} \\ {PB = 1  x.} \end{array}\)
In particular,
\(y = \ln \dfrac{AP \cdot BO}{PB \cdot OA} = \ln \dfrac{1 + x}{1  x}.\)
Taking the exponential function of the left and the right hand side and applying obvious algebra manipulations, we get that
\(x=\dfrac{e^y1}{e^y+1}\).
Lemma \(\PageIndex{3}\)
Assume the points \(P\), \(Q\), and \(R\) appear on one hline in the same order. Then
\(PQ_h + QR_h = PR_h.\)
 Proof

Note that
\(PQ_h + QR_h = PR_h\)
is equivalent to
\[\delta (P, Q) \cdot \delta (Q,R) = \delta (P, R).\]
Let \(A\) and \(B\) be the ideal points of \((PQ)_h\). Without loss of generality, we can assume that the points \(A\), \(P\), \(Q\), \(R\), and \(B\) appear in the same order on the circline containing \((PQ)_h\). Then
\(\begin{array} {rcl} {\delta (P, Q) \cdot \delta (Q, R)} & = & {\dfrac{AQ \cdot BP}{QB \cdot PA} \cdot \dfrac{AR \cdot BQ}{RB \cdot QA} =} \\ {} & = & {\dfrac{AR \cdot BP}{RB \cdot PA} =} \\ {} & = & {\delta (P, R).} \end{array}\)
Hence 12.3.1 follows.
Let \(P\) be an hpoint and \(\rho>0\). The set of all hpoints \(Q\) such that \(PQ_h=\rho\) is called an hcircle with the center \(P\) and the hradius \(\rho\).
Lemma \(\PageIndex{4}\)
Any hcircle is a Euclidean circle that lies completely in the hplane.
More precisely for any hpoint \(P\) and \(\rho\ge 0\) there is a \(\hat\rho\ge 0\) and a point \(\hat P\) such that
\(PQ_h= \rho \Leftrightarrow \hat{P}Q= \hat{\rho}\)
for any hpoint \(Q\).
Moreover, if \(O\) is the center of the absolute, then

\(\hat{O}=O\) for any \(\rho\) and

\(\hat{P} \in (OP)\) for any \(P\ne O\).
 Proof

According to Lemma \(\PageIndex{2}\), \(OQ_h\z= \rho\) if and only if
\(OQ= \hat{\rho}=\dfrac{e^{\rho}1}{e^{\rho}+1}.\)
Therefore, the locus of hpoints \(Q\) such that \(OQ_h= \rho\) is a Euclidean circle, denote it by \(\Delta_{\rho}\).
If \(P \ne O\), then by Lemma \(\PageIndex{1}\) and the main observation (Theorem \(\PageIndex{1}\)) there is inversion that respects all hnotions and sends \(O \mapsto P\).
Let \(\Delta_{\rho}'\) be the inverse of \(\Delta_{\rho}\). Since the inversion preserves the hdistance, \(PQ_h=\rho\) if and only if \(Q \in \Delta_{\rho}'\).
According to Theorem 10.3.1, \(\Delta_\rho'\) is a Euclidean circle. Let \(\hat P\) and \(\hat\rho\) denote the Euclidean center and radius of \(\Delta_\rho'\).
Finally, note that \(\Delta_\rho'\) reflects to itself across \((OP)\); that is, the center \(\hat P\) lies on \((OP)\).
Exercise \(\PageIndex{1}\)
Assume \(P\), \(\hat P\), and \(O\) are as in the Lemma \(\PageIndex{1}\) and \(P \ne O\). Show that \(\hat{P} \in [OP]\).
 Hint

Let \(X\) and \(Y\) denote the point of intersections of \((OP)\) and \(\Delta_{rho}'\). Consider an isometry \((OP) \to \mathbb{R}\) such that \(O\) corresponds to 0. Let \(x, y, p\), and \(\hat{p}\) denote the real number corresponding to \(X, Y, P\), and \(\hat{P}\).
We can assume that \(p > 0\) and \(x < y\). Note that \(\hat{p} = \dfrac{x + y}{2}\) and
\(\dfrac{(1 + x) \cdot (1  p)}{(1  x) \cdot (1 + p)} = \dfrac{(1 + p) \dot (1  y)}{(1  p) \cdot (1 + y)}\).
It remains to show that all this implies \(0 < \hat{p} < p\).