Non-Euclidean Geometry

Inversion in Circle

Hyperbolic geometry can be modelled by the Poincaré disc model or the Poincaré halfplane model. In both those models circle inversion is used as reflection in a geodesic.

In Euclidean geometry a triangle that is reflected in a line is congruent to the original triangle. Both distances and angles are preserved when reflecting in a line.

Using the Poincaré disc model or the Poincaré halfplane model, distances and angles are preserved when objects are reflected in a geodesic. This is achieved by using circle inversion for reflections. Hyperbolic angles are the same as Euclidean angles (measured by using tangents to geodesics). Hyperbolic distance is defined in such a way that it is preserved when reflecting in a geodesic. The hyperbolic distance between two points is not the same as the Euclidean distance.

By using inversion in circle as a hyperbolic version of the Euclidean reflection in a line, we will be able to construct hyperbolic tools such as midpoint, perpendicular line and perpendicular bisector. We will start by defining circle inversion and then show all the properties of circle inversion that are needed to construct hyperbolic tools.

Definition of circle inversion

If the inversion circle $$c$$ has radius $$r$$ and centre $$O$$, then the inverted point $$A'$$ of a point $$A$$ lies on a ray from $$O$$ to $$A$$, and

\begin{equation*} OA\cdot {OA}' = r^2. \end{equation*}

As the point $$A$$ tends to $$O$$, the inverted point $$A'$$ tends to infinity. The inversion of $$O$$ is not defined. Points on the circle $$c$$ are inverted to themselves. Points inside $$c$$ are inverted to points on the outside, and vice versa. Inversion in a circle is a transformation that flips the circle inside out.

It is possible to construct the inverted point using a ruler and compass. In GeoGebra you can use the tool Reflect Object in Circle to create the inverted point.

In order to show the properties of circle inversion that are needed, we need to find relations between points and inverted points.

Theorem 1

Let $$O$$ be the centre of an inversion circle $$c$$. Let $$A$$ and $$B$$ be distinct points inside the circle not equal to $$O$$ and such the $$A$$, $$B$$, and $$O$$ are not collinear. Let $$A'$$ and $$B'$$ be the inverted points. Then $$\bigtriangleup OAB \sim \bigtriangleup OB'A'$$.

Proof

The proof is the task of Exercise 1.

Generalized circles inverted in a circle

When a circle is inverted in a circle it is inverted to either a circle or a line. The same holds when a line is inverted in a circle, it is inverted to either a circle or a line.

It is convenient not to distinguish between a circle and a line, and instead use a so-called generalized circle, i.e. a geometric object that is either a circle or a line. Using GeoGebra, you can use the tool Reflect Object in Circle on circles and lines.

The different cases for circle inversion of circles and lines are as follows:

Theorem 2

Let $$O$$ be the centre of an inversion circle $$c$$.

1. A line through $$O$$ is inverted to itself.
2. A line not through $$O$$ is inverted to a circle through $$O$$. Conversely a circle through $$O$$ is inverted to a line not through $$O$$.
3. A circle not through $$O$$ is inverted to a circle not through $$O$$.

Proof

We must prove the theorem for each of the three cases.

Case 1 ‐ A line through $$O$$ is inverted to itself.

Let $$l$$ be a line through $$O$$ and let $$A$$ and $$B$$ be two points on $$l$$. The inverted line is defined by the inverted points $$A'$$ and $$B'$$. The inverted points are on rays from $$O$$ to $$A$$ and $$B$$ respectively. Since both rays are on $$l$$ the inverted points must both be on $$l$$. Hence the inverted line coincides with $$l$$.

Case 2 ‐ A line not through $$O$$ is inverted to a circle through $$O$$. Conversely a circle through $$O$$ is inverted to a line not through $$O$$.

The proof is the task of Exercise 3. The task is given with hints.

Case 3 ‐ A circle not through $$O$$ is inverted to a circle not through $$O$$.

The proof is the task of Exercise 4. The task is given with hints.

Angle preservation

Using generalized circles and circle inversion, we can form an angle in three different ways:

• an angle between two circles,
• an angle between two lines,
• an angle between a circle and a line.

An angle at a point on a circle is measured from the tangent through that point. There are always two supplementary angles between two generalized lines.

If you reflect a triangle in a line, the orientation of the points is reversed.

When you invert three points in a circle, their orientation is reversed, just as when reflecting in a line.

Theorem 3

Angles are preserved when generalized circles are inverted in a circle.

Proof

We will not prove the theorem for all cases but sketch out the proof for the case where the angle is formed by two circles.

Let $$O$$ be the centre of the inversion circle and let $$r$$ be a ray from $$O$$ that intersects both circles. Let $$A$$ be an intersection point between $$r$$ and one of the circles and let $$A'$$ be the inverted point. Let $$B$$ be an intersection point between $$r$$ and the other circle and let $$B'$$ be the inverted point. Let $$C$$ be an intersection point between the circles and let $$C'$$ be the inverted point.

The angle between two circles is the angle between the tangent lines of the circles at $$C$$. Instead of using the angle between tangent lines, we will use the angles $$\angle ACB$$ and $$\angle A'C'B'$$. As the ray is moved so $$A$$ and $$B$$ tends to $$C$$, the angle $$\angle ACB$$ will tend to the angle between the tangents of the circles at $$C$$. The angle $$\angle A'C'B'$$ will tend to the angle between the tangents of the inverted circles at $$C'$$

Now it suffices to show that $$\angle ACB = \angle A'C'B'$$. This is shown by using the same steps as in Exercise 4.

Other cases of angles between circles or lines can be proved in a similar way.

Note that when two circles are reflected in a circle, the angles are preserved but since the orientation of points is reversed the orientation of angles is also reversed. When forming the angles between two circles you can always do it so that exactly one of the supplementary angles is inside both circles.

After inversion the two angles are reversed with regards to which angle that is inside both circles.

The preservation of angles also holds when lines are inverted to circles and vice versa.

As a consequence of angle preservation, objects that are tangent to each other, are inverted to objects that are also tangent to each other.

Perpendicular circles

The Poincaré disc is the set of points inside a circle. When using the Poincaré disc model, only points in the Poincaré disc are considered. The boundary of the disc is the circle at infinity, $$C_\infty$$.

A geodesic through points $$A$$ and $$B$$ is defined as the circular arc through $$A$$ and $$B$$ that is perpendicular to $$C_\infty$$. If $$A$$ and $$B$$ lie on the diameter of $$C_\infty$$, that diameter is the geodesic through $$A$$ and $$B$$. A geodesic is the hyperbolic equivalence of a Euclidean line.

The reason why perpendicular generalized arcs are used as hyperbolic lines to model hyperbolic geometry, is because of the properties of perpendicular circles when doing circle inversion.

Theorem 4

If a circle $$d$$ is inverted in another circle $$c$$, then $$d$$ is inverted to itself if and only if $$c$$ and $$d$$ are perpendicular.

Proof

To begin with we assume that $$c$$ and $$d$$ are perpendicular. We will show that $$d$$ is inverted to itself. For this part of the proof we will use that a tangent to a circle through a point is perpendicular to the radius through the same point.

Let $$A$$ and $$B$$ be the intersection points of $$c$$ and $$d$$. The lines through $$A$$ and $$B$$ tangent to $$d$$ must go through the centre of $$c$$. The lines through $$A$$ and $$B$$ tangent to $$c$$ must go through the centre of $$d$$. Since a circle is uniquely defined by its centre and a point on a circle there can only be one circle through $$A$$ and $$B$$ that is perpendicular to $$c$$.

When $$d$$ is inverted in $$c$$ we know from Theorem 3 that the inverted circle is perpendicular to $$c$$. We can tell from the definition of circle inversion that $$A$$ and $$B$$ are both inverted to themselves. Therefore the inverted circle is a circle through $$A$$ and $$B$$ perpendicular to $$c$$ and since such circles are uniquely defined, the inverted circle must be the same as $$d$$.

Next we show that if $$d$$ is not perpendicular to $$c$$, the inversion of $$d$$ is not the same circle as $$d$$.

For a circle not perpendicular to the inversion circle, the inverted circle will still have the same angles to the inversion circle, but the orientation of the angles is reversed, for that reason the inverted circle cannot be the same circle as $$d$$.

Geodesics in the Poincaré disc

Consider a geodesic in the Poincaré disc. The geodesic is perpendicular to $$C_\infty$$ and hence by Theorem 4 we know that $$C_\infty$$ is inverted to itself when inverted in the geodesic. If a point inside $$C_\infty$$ is inverted in a geodesic, the inverted point will also be on the inside of $$C_\infty$$, on the other side of the geodesic.

Theorem 5

Let $$C_\infty$$ be the boundary of the Poincaré disc. Let $$c$$ be a circle that intersects $$C_\infty$$. Let $$A$$ be a point on $$c$$ and $$A'$$ the inversion of $$A$$ in $$C_\infty$$. Then $$A'$$ is a point on $$c$$ if and only if $$c$$ is perpendicular to $$C_\infty$$.

Proof

We start by assuming that $$c$$ is perpendicular to $$C_\infty$$. We must show that in this case $$A'$$ must be a point on $$c$$. From Theorem 4 we know that $$c$$ is inverted to itself and hence $$A'$$ must also be a point on $$c$$.

Next we assume that $$A'$$ is a point on $$c$$. We must show that in this case $$c$$ must be perpendicular to $$C_\infty$$.

We make following construction: Let $$M$$ be the centre of circle $$c$$. Mark one of the intersection points between the two circles and call it $$P$$. Make a ray through $$O$$ and $$A$$. Make a line through $$M$$ that is perpendicular to the ray and mark the intersection point $$B$$.

To show that $$c$$ is perpendicular to $$C_\infty$$, we must show that $$\angle OPM$$ is a right angle. By using the converse of Pythagoras' theorem, it suffices to show that $${OP}^2+{MP}^2={OM}^2$$. We have that

\begin{align*} {OP}^2 & = OA\cdot {OA}'\\ & = (OB - AB)(OB + AB) \\ & = {OB}^2-{AB}^2 \\ & = ({OM}^2-{MB}^2)-{AB}^2 \\ & = {OM}^2-({MB}^2+{AB}^2) \\ & ={OM}^2-{MA}^2 \\ & = {OM}^2-{MP}^2. \end{align*}

Hence, any circle through a point $$A$$ and its inverted point $$A'$$ is perpendicular to the inversion circle.

A special case of a perpendicular circle through $$A$$ and $$A'$$ is shown below. Let $$M$$ be the midpoint of $$A$$ and $$A'$$, and let $$c$$ be the circle through $$A$$ that has $$M$$ as its centre. Then $$c$$ is perpendicular to $$C_\infty$$ and also perpendicular to the line through $$O$$ and $$A$$.

Hyperbolic distance

When inverting in a circle, angles are preserved but distances are not. There is, however, a relation between distances that is preserved even after a circle inversion.

Definition of the cross ratio

Given four points $$A$$, $$B$$, $$C$$, $$D$$, the cross ratio $$(A,B; C, D)$$ is defined by

$(A,B; C, D) = \frac{AC \cdot BD }{BC\cdot AD}.$

We will show that the cross ratio is preserved if four distinct points are inverted in a circle.

Theorem 6

The cross ratio is invariant under circle inversion.

Proof

We will show that if four distinct points $$A$$, $$B$$, $$C$$, $$D$$ are inverted in a circle to the points $$A'$$, $$B'$$, $$C'$$, $$D'$$, then

$(A,B; C, D) = (A',B'; C', D') \iff \frac{AC \cdot BD }{BC\cdot AD} = \frac{A'C' \cdot B'D' }{B'C'\cdot A'D'}.$

We know from Theorem 1 that for any pair of inverted points $$A$$ and $$B$$ that are not collinear with $$O$$, $$\Delta OAB\ \sim \Delta OB'A'$$. Hence

$\label{eq:oaob} \frac{AB}{A'B'} = \frac{OA}{{OB}'}.$

Next consider the case where $$A$$, $$B$$, $$O$$ are collinear, as in the figure below.

With this arrangement we get that

$\frac{AB}{OA} = \frac{OA+OB}{OA} = 1+\frac{OB}{OA} = 1+\frac{r^2/{OB}'}{r^2/{OA}'} = 1+\frac{{OA}'}{{OB}'} = \frac{{OB}'+{OA}'}{{OB}'} = \frac{A'B'}{{OB}'}.$

If $$A$$ and $$B$$ are on the same ray from $$O$$ (not on opposite sides as in the figure above), a similar calculation will yield the same result.

The equality can be applied to all points defining the cross ratio, hence

$\frac{AC}{A'C'}=\frac{OA}{{OC}'}, \hspace{1cm} \frac{BC}{B'C'} = \frac{OB}{{OC}'}, \hspace{1cm} \frac{BD}{B'D'} = \frac{OB}{{OD}'}, \hspace{1cm} \frac{AD}{A'D'} = \frac{OD}{{OA}'}.$

Multiplying and using these relations we get that

$\frac{AC}{A'C'}\cdot \frac{B'C'}{BC}\cdot \frac{BD}{B'D'}\cdot \frac{A'D'}{AD} = \frac{OA}{{OC}'}\cdot \frac{{OC}'}{OB}\cdot \frac{OB}{{OD}'}\cdot \frac{{OD}'}{OA}=1.$

We're not interested in distances to $$O$$ but we have shown that

$\frac{AC}{A'C'}\cdot \frac{B'C'}{BC}\cdot \frac{BD}{B'D'}\cdot \frac{A'D'}{AD} =1,$

which after rearrangement shows that the cross ratio is preserved when inverting in a circle.

We will use the cross ratio to define hyperbolic distance.

Definition of hyperbolic distance

Let $$A$$ and $$B$$ be points in the Poincaré disc. Let $$P$$ and $$Q$$ be endpoints of the hyperbolic line through $$A$$ and $$B$$. Then the hyperbolic distance $$d(A,B)$$ between $$A$$ and $$B$$ is defined by:

$d(A,B) = |\ln(A,B; P,Q)|=\left|\ln\left(\frac{AP \cdot BQ}{BP \cdot AQ}\right)\right|$

$$P$$ and $$Q$$ are so-called ideal points. Since the circle itself lies at infinity, $$P$$ and $$Q$$ do not belong to our hyperbolic universe. Since $$A$$ and $$B$$ can never be equal to either $$P$$ or $$Q$$, the distance is always defined.

In order for the hyperbolic distance to have the same functionality as Euclidean distance, it must have a number of properties.

Theorem 7

For all points $$A$$, $$B$$, $$C$$ in the Poincaré disc, following statements are true:

• $$d(A,B)\geq 0$$
• $$d(A,A)= 0$$
• $$A\neq B \Rightarrow d(A,B)\neq 0$$
• $$d(A,B) = d(B,A)$$
• If $$A$$ tends to infinity then $$d(A,B)$$ tends to infinity.
• If $$C$$ lies between $$A$$ and $$B$$ on the geodesic through $$A$$ and $$B$$, then
$$d(A,C)+d(C,B) = d(A, B)$$.

Proof

We will use the definition of hyperbolic distance

$d(A,B) = |\ln(A,B; P,Q)|=\left|\ln\left(\frac{AP \cdot BQ}{BP \cdot AQ}\right)\right|.$
• $$d(A,B)\geq 0$$: This follows from the directly from the definition.
• $$d(A,A)=\left|\ln\left(\frac{AP \cdot AQ}{AP \cdot AQ}\right)\right| = \ln(1) =0$$
• $$A\neq B \Rightarrow d(A,B)\neq 0$$:
If $$AP > BP$$ then $$AQ < BQ \implies AP\cdot BQ > BP \cdot AQ$$, hence $$\frac{AP \cdot BQ}{BP \cdot AQ} > 1 \implies d(A,B) > 0$$.
If $$AP < BP$$ the same line of reasoning can be used.
• $$d(A,B) = \left|\ln\left(\frac{AP \cdot BQ}{BP \cdot AQ}\right)\right| = \left|-\ln\left(\frac{AP \cdot BQ}{BP \cdot AQ}\right)\right| = \left|\ln\left(\frac{BP \cdot AQ}{AP \cdot BQ}\right)\right| = d(B,A)$$
• If $$A$$ tends to $$P$$ (or $$Q$$), $$AP$$ (or $$AQ$$) tends do 0 and then $$d(A,B)$$ tends to infinity.
• If $$C$$ lies between $$A$$ and $$B$$ on the geodesic through $$A$$ and $$B$$, then

$$d(A,C)+d(C,B) = \left|\ln\left(\frac{AP \cdot CQ}{CP \cdot AQ}\right)\right| + \left|\ln\left(\frac{CP \cdot BQ}{BP \cdot CQ}\right)\right|$$

If $$AQ < AP$$ then
$$AQ < CQ < BQ$$ and $$BP < CP < AP$$, hence
$$AP\cdot CQ > CP\cdot AQ$$ and $$CP \cdot BQ > BP \cdot CQ$$,
hence it is safe to remove the absolute signs.

$$d(A,C)+d(C,B) = \ln\left(\dfrac{AP \cdot CQ \cdot CP \cdot BQ}{CP \cdot AQ \cdot BP \cdot CQ}\right) = \ln\left(\dfrac{AP \cdot BQ}{AQ \cdot BP}\right) = d(A,B)$$.
The case when $$AQ > AP$$ is shown using the same line of reasoning.

In Theorem 6 we showed that the cross ratio is invariant under circle inversion for four distinct points $$A$$, $$B$$, $$C$$ and $$D$$. We will use this to show that the hyperbolic distance between two points $$A$$ and $$B$$ is preserved when inverting the points in a geodesic.

Theorem 8

Inversion in a geodesic preserve distances

Proof

Let $$C_\infty$$ be the boundary of a Poincaré disc. Let $$g$$ be a geodesic in $$C_\infty$$ and let two points $$A$$ and $$B$$ in $$C_\infty$$ be inverted in $$g$$ to points $$A'$$ and $$B'$$.

We must show that $$d(A,B) = d(A', B')$$.

Let $$P$$ and $$Q$$ be the endpoints of the geodesic through $$A$$ and $$B$$. Let $$P'$$ and $$Q'$$ be the endpoints of the geodesic through $$A'$$ and $$B'$$. We know that points on $$C_\infty$$ are inverted to points on $$C_\infty$$. For that reason it must be that $$P$$ is inverted to $$P'$$ and $$Q$$ is inverted to $$Q'$$.

Since the cross ratio is preserved under circle inversion, so is the distance between $$A$$ and $$B$$.

Summary

In summary we have shown that both angles and hyperbolic distances are preserved when inverting in a geodesic in the Poincaré disc, hence circle inversion in a geodesic is the hyperbolic version of reflection in a line.

Exercises

Exercise 1

Prove Theorem 1, i.e.:

Let $$O$$ be the centre of an inversion circle $$c$$. Let $$A$$ and $$B$$ be distinct points inside the circle not equal to $$O$$ and such the $$A$$, $$B$$, and $$O$$ are not collinear. Let $$A'$$ and $$B'$$ be the inverted points, then $$\bigtriangleup OAB \sim \bigtriangleup OB'A'$$.

Exercise 2

Thales theorem states that if two vertices of a triangle are the endpoints of a diameter of a circle, and if the third vertex also lies on the circle, then the angle at the third vertex is a right angle.

The converse of Thales theorem is: for any triangle $$\bigtriangleup ABC$$ such that $$\angle B$$ is a right angle, a circle can be constructed such that all vertices of the triangle are on the circle and $$A$$ and $$B$$ are on a diameter of the circle.

1. Prove Thales theorem.
2. Prove the converse of Thales theorem.

Exercise 3

Let $$O$$ be the centre of an inversion circle. A line not through $$O$$ is inverted to a circle through $$O$$. Conversely a circle through $$O$$ is inverted to a line not through $$O$$.

Let $$l$$ be a line not through $$O$$. Make another line through $$O$$ perpendicular to $$l$$ and let $$A$$ be the intersection between the two lines. Let $$B$$ be another point on $$l$$.

1. Show that $$\angle OB'A'$$ is a right angle.
2. Use the conclusion from a) and the conclusion from Exercise 2 to prove that a line not through $$O$$ is inverted to a circle through $$O$$.

By reversing the process, we can show that circles through $$O$$ (excluding the point $$O$$) are inverted to lines not through $$O$$. You don't have to do this.

Exercise 4

Let $$O$$ be the centre of an inversion circle. A circle not through $$O$$ is inverted to a circle not through $$O$$.

You only need to show this for a circle that is inside the circle of inversion.

Let $$d$$ be a circle not through $$O$$ inside the circle of inversion. Let $$A$$ be the point on $$d$$ closest to $$O$$ and let $$B$$ be the point on $$d$$ furthest away from $$O$$. Let $$C$$ be any point on $$d$$ distinct from $$A$$ and $$B$$. Let $$A'$$, $$B'$$ and $$C'$$ be the inversions of the points $$A$$, $$B$$ and $$C$$ respectively.

1. Show that $$\angle OAC = \angle OC'A'$$ and that $$\angle OBC = \angle OC'B'$$.
2. Show that $$\angle ACB$$ is a right angle.
3. Show that $$\angle B'C'A'$$ is a right angle.
4. Show that $$d$$ is inverted to the circle through $$A'$$, $$B'$$ and $$C'$$.

Exercise 5

GeoGebra construction of Ford circles

Start by watching Francis Bonahan talk about Funny Fractions and Ford Circles at Numberphile's YouTube channel.

1. Use any GeoGebra tools to construct three circles along the $$x$$-axis as shown below.
2. Mark the intersection points shown below. The $$x$$-axis can be seen as a generalized circle that is a line. Construct the two new blue circles by using the tool Circle through 3 points.
3. Construct the two new red circles shown below by using the tool Reflect about circle. You are not alllowed to use any other tool.
4. Explain why your construction in c) works.

The procedure can be repeated.

references:

Marvin Jay Greenberg - Euclidean and Non-Euclidean Geometries, Development and History