Inversion in Circle

In order to understand the constructions of the hyperbolic tools on the next page, it is necessary to know about some properties of circle inversion.

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 $$O$$ and $$A$$, and

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

As a point $$A$$ gets closer and closer to $$O$$, the inverted point $$A'$$ tends to infinity. The inversion of $$O$$ is not defined unless a point at infinity is introduced. 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. For more information about the construction of an inverted point, and an interactive inverted paint example, see Paint Circle-Inverted Mondrian!

Similar triangles

Let $$O$$ be the centre of the inversion circle. Let $$A$$ and $$B$$ be distinct points not equal to $$O$$. Consider the triangles $$\Delta OAB$$ and $$\Delta {OB}'A'$$.

We know that $$OA\cdot {OA}' = OB\cdot {OB}' = r^2$$. Since $$\dfrac{OA}{OB} = \dfrac{OA'}{OB'}$$, and $$\angle AOB = \angle A'{OB}'$$, the SAS-criterion for similar triangles holds. Hence, the triangles are similar.

$$\label{eq:similar} \Delta OAB = \Delta {OB}'A'$$

Generalized circles and angle preservation

Let $$O$$ be the centre of the inversion circle. A circle through $$O$$ is inverted to a line. Conversely a line is inverted to a circle through $$O$$. A circle that does not contain $$O$$ is inverted to a circle. It is convenient not to distinguish between circles and lines, and instead use so called generalized circles. GeoGebra can handle generalized circles. Try it out by using the tool Reflect Object in Circle on circles and lines!

Angles are preserved when points are inverted in a circle. As a consequence, objects that are tangent to each other, are inverted to objects that are also tangent to each other. For an example where the preservation of tangency is used, see Geometry - Apollonian gasket.

In order to show that generalized circles are inverted to generalized circles, all possible cases must be considered.

Lines and circles through the centre of the inversion circle

If two points $$A$$ and $$B$$ are both on a line $$l$$ through the centre $$O$$; then both rays from $$O$$ to $$A$$, and from $$O$$ to $$B$$ are on $$l$$. Hence the inverted points $$A'$$ and $$B'$$ are also both on $$l$$. Lines through the centre $$O$$ are mapped onto themselves when inverted in a circle.

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

Since $$\Delta OAB \sim \Delta OB'A'$$, we must have that $$\angle OB'A' = \angle OAB = 90^\circ$$. Using Thales theorem and the converse of Thales theorem, we see that $$B'$$ is on a circle through $$A$$ and $$A'$$. We have shown that lines not through $$O$$ are inverted to circles through $$O$$ (excluding the point $$O$$). By reversing the process, we can show that circles through $$O$$ are inverted to lines not through $$O$$.

Circles that are not through the centre of the inversion circle

Let $$c$$ be a circle not through $$O$$. Let $$A$$ be the point on $$c$$ closest to $$O$$ and let $$B$$ be the point on $$c$$ furthest away from $$O$$. Let $$C$$ be any point on $$c$$ distinct from $$A$$ and $$B$$. The figure below shows the case when $$c$$ is inside the circle of inversion.

Using similarity, Thales theorem, and the converse of Thale's theorem; it is possible to show that $$C'$$ is on a circle through $$A'$$ and $$B'$$. The proof is sketched in the figure above but the details are omitted. Other placements of $$c$$ can be sketched in an analogous way. In summary: circles not through $$O$$ are inverted to circles.

Angles are preserved

The angle between two circles is the angle between the tangent lines of the circles at an intersection point. Instead of using the angle between tangent lines, we will use the angle $$\angle ACB$$ where $$C$$ is an intersection point, and the points $$A$$ and $$B$$ are on a line and also on one of the circles; as in the figure below.

The angle between the tangent lines of the circles, is the limit of the angle $$\angle ACB$$ as the line through $$A$$ and $$B$$ gets closer and closer to the line through $$O$$ and $$C$$. It is possible to show that $$\angle ACB = \angle B'C'A$$. The proof is sketched in the figure above but the details are omitted. Other cases of angles between circles or lines can be proved in an analogous way. In summary, the angle between two generalized circles is preserved when the generalized circles are inverted in a circle.

Circle inversion when the circles are perpendicular to each other

The Poincaré disc (in 2D) is the area inside a circle. The circle is not included. When using the Poincaré disc model, only points in the Poincaré disc are considered. The Poincaré disc makes up the entire world. 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. A geodesic is a hyperbolic 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.

Circles perpendicular to the circle of inversion are inverted to themselves

Given a circle $$c$$ and two distinct points on $$c$$; then there is a unique circle through the points that is perpendicular to $$c$$. The proof is omitted but sketched in the worksheet below.

If the red circle is inverted in the gray circle, the big red points are inverted to themselves, and the inverted circle is also perpendicular to the gray circle (since angles are preserved). Since there is only one unique circle through the points, perpendicular to the gray circle, the inverted circle must be the same as the original circle. A circle is inverted to itself, if and only if the circle is perpendicular to the circle of inversion.

For a circle not perpendicular to the the inversion circle, the inverted circle will still have the same angles to the inversion circle. Since there are two angles between two lines, and since the orientation of points is changed when inverting in a circle; the inverted circle will not be the same as the original circle. A circle is inverted to itself, if and only if the circle is perpendicular to the circle of inversion.

Consider a geodesic on the Poincaré disc. The geodesic is perpendicular to $$C_\infty$$, so $$C_\infty$$ 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.

Perpendicular circles and points of inversion

Let $$A$$ be a point, and $$c$$ a circle through $$A$$ that is perpendicular to $$C_\infty$$. If $$A$$ is inverted in $$C_\infty$$ the inverted point $$A'$$ must also be on $$c$$ (since $$c$$ is inverted to itself).

The converse statement is: Let $$A$$ be a point on a circle $$c$$. Let $$A'$$ be the inversion of $$A$$ in $$C_\infty$$. If $$A'$$ also lies on $$c$$, $$c$$ must be perpendicular to $$C_\infty$$.

To show the converse statement make following construction: Let $$M$$ be the center 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 $$B$$. As in the figure below.

To show that $$c$$ is perpendicular to $$C_\infty$$, we must show that $$\angle OPM = 90^\circ$$. 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 $$A$$.

Definition of the cross ratio

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

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

The cross ratio is invariant under circle inversion

If the four points are inverted in a circle to the points $$A', B', C', D'$$, then

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

We know from equation $$\eqref{eq:similar}$$ 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.

In this case we get

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

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. For all points $$A$$ and $$B$$ that are distinct from the center $$O$$, equation $$\eqref{eq:oaob}$$ must be true.

Equation $$\eqref{eq:oaob}$$ can be applied to all points defining the cross ratio:

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

Multiplying and using these relations we get that

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

Which proves statement $$\eqref{eq:crossratio}$$, that the cross ratio is preserved when inverting in a circle.

Hyperbolic distance

Let $$A$$ and $$B$$ be points on 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 is 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.

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

• $$d(A,B)\geq 0$$
• $$d(A,A)=\left|\ln\left(\frac{AP \cdot AQ}{AP \cdot AQ}\right)\right| = \ln(1) =0$$
• $$A\neq B \Rightarrow d(A,B)\neq 0$$
• $$d(A,B) = \left|\ln\left(\frac{AP \cdot BQ}{BP \cdot AQ}\right)\right| = \left|-\ln\left(\frac{AP \cdot BQ}{BP \cdot AQ}\right)\right| = \left|\ln\left(\frac{BP \cdot AQ}{AP \cdot BQ}\right)\right| = d(B,A)$$
• As $$A$$ gets closer and closer to $$P$$ (or $$Q$$), $$AP$$ (or $$AQ$$) tends do 0 and $$d(A,B)$$ 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 the ordering of the points is as in the figure to the right, then $$AP \cdot CQ > CP \cdot AQ$$ and $$CP \cdot BQ > BP \cdot CQ$$, hence it is safe to remove the absolute signs.

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

Circle inversion in a geodesic is the hyperbolic version of reflection in a line

Let $$C_\infty$$, with centre $$O$$, be the boundary of a Poincaré disc. Let two points $$A$$ and $$B$$ be inverted in a geodesic $$g$$ to the points $$A'$$ and $$B'$$. We want to show that the hyperbolic distance is preserved, i.e that $$d(A,B) = d(A', B')$$. Since the hyperbolic distance depends on the endpoints of the geodesic through $$A$$ and $$B$$, we must first consider the endpoints under inversion in $$g$$.

The endpoints of a geodesic are inverted to the endpoints of the inverted geodesic

Let $$P$$ and $$Q$$ be points on $$C_\infty$$ that are inverted in the geodesic $$g$$ to points $$P'$$ and $$Q'$$. By using the intersection of tangents, it is possible to construct a circle through $$P$$ and $$P'$$ that is perpendicular to $$C_\infty$$. In the same way we can construct a circle through $$Q$$ and $$Q'$$ that is perpendicular to $$C_\infty$$. Those two circles are blue in the worksheet below. We can also construct a circle through $$P'$$ and $$Q'$$ that is perpendicular to $$C_\infty$$, the yellow circle in the worksheet below.

A geodesic through $$P$$ and $$Q$$ is an arc on the red circle through $$P$$ and $$Q$$. If this geodesic is inverted in $$g$$, the inverted geodesic must be perpendicular to $$C_\infty$$, and it must also be tangent to the blue circles, since tangency is preserved under circle inversion. Hence, the inverted geodesic must be an arc on the yellow circle. The endpoints of the inverted geodesic are the intersection points with the blue circles, and those points are $$P'$$ and $$Q'$$.

Inversion in a geodesic preserves distances

Let $$A$$ and $$B$$ be distinct points on the Poincaré disc, and let $$P$$ and $$Q$$ be the endpoints of the geodesic through $$A$$ and $$B$$. Let $$A$$ and $$B$$ be inverted in a geodesic to $$A'$$ and $$B'$$. The endpoints of the geodesic through $$A'$$ and $$B'$$ are the inverted points $$P'$$ and $$Q'$$. Since the cross ratio is preserved under circle inversion, so is the distance between $$A$$ and $$B$$.

$$(A, B; P, Q) = (A',B'; P', Q') \Longrightarrow d(A, B) = d(A', B')$$

references:

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