# Inversive Plane Geometry

An inversive plane is a geometry with three undefined notions: points, circles, and an incidence relation between points and circles, satisfying the following three axioms:

(I.1) Through any three distinct points there is

exactly one circle.

(I.2) If P and Q are points, and C is a circle

passing through P but not Q, then there

is a unique circle C ′ passing through Q

such that C ∩ C′ = {P}.

(I.3) There exist four points which do not lie on a common circle .

A model of these axioms is provided by the points and circles lying on a
sphere S in Euclidean 3-space. Another (obtained by stereographically projecting
S onto a plane) is the extended Euclidean

plane E = R^{2} ∪ {∞} consisting of the Euclidean plane R^{2} together with one new
point ∞ called the

point at infinity. (This is different from the real projective plane in which
there are many points at

infinity.) The circles of E are of two types : the ordinary circles of R^{2}, and
sets of the form l ∪ {∞}

where l is a line of R ^{2}. Because the second model is obtained from the first by
stereographic

projection, the two models are isomorphic. Other models exist (in particular
finite models) but we

will be primarily concerned with the model E described above, called the real
inversive plane. In this case we may reasonably measure distances and angles.

**Straightedge and Compass Constructions**

It is often instructive to provide, along with the relevant definitions,
straightedge-and-compass constructions; and we shall often do so when this is
feasible. Recall that the following procedures can be implemented using
straightedge and compass:

1. Given a point P and a line l , construct the line through P perpendicular to
l.

2. Find the midpoint of a line segment AB.

3. Bisect an angle ABC.

Using these basic constructions we can perform others, for example:

**Lemma 1**. Given line segments of lengths a and b, one may construct a
line segment of length
Thus given a rectangle, we may construct a square with the same area.

Proof. Construct a line segment AB containing a point C such that AC = a and BC = b. Construct the midpoint O of AB. Construct a semicircle centered at O with radius OA = OB. Construct a perpendicular l to AB at C. Let D be the point of intersection of l with the semicircle. We will show that CD has the required length Observe that triangles ACD and DCB are similar since corresponding angles are equal. Therefore

i.e. CD^{2} = AC · CB = ab as required.

We will show that given a circle C and a point P outside C, one may construct a tangent from P to C. This construction relies on the following result.

Lemma 2. Let C be a circle and let P be a point outside C. Consider a

tangent PT to C, and a secant through P meeting C at A and A′. Then

PA · PA′ = PT^{2}.

Proof. Let O be the center of C, and let r be the radius. Drop a perpendicular OB from O to the secant, as shown. Then

by Pythagoras’ Theorem for triangle PTO | ||

since OA = OT = r | ||

by Pythagoras’ Theorem for triangle ABO | ||

by Pythagoras’ Theorem for triangle PBO | ||

= (PB + AB) (PB – AB) | ||

= PA′ · PA | since BA′ = BA. |

**Lemma 3.** Given a circle C and a point P outside C, one may construct
tangents from P to C.

Proof. Construct any secant to C through P, and let A and A′ be the points of intersection of this secant with C. By Lemma 1 one may construct a line segment of length which is the length of the required tangent. Set the radius of the compass to this length and draw an arc centered at P to intersect C at two points Q and R. Then PQ and PR are the required tangents.

**The Real Inversive Plane**

Two circles in E are orthogonal (i.e. perpendicular) if they intersect at right angles. Note that in this case the circles meet twice, and if the angle at one point of intersection is 90°, the angle at the other point of intersection must also be 90°.

Two orthogonal circles

Given a circle C with center O, and a point P, we define the inverse P′ of P in C as follows. The inverse of every point of C is itself (P′ = P). The inverse of O is ∞, and the inverse of ∞ is O. If P is inside C (but different from O ) then extend the line OP beyond the circle C and erect a perpendicular to this line at P. This perpendicular meets C at points Q and R, say. The tangents to C at Q and at R meet at P′. (Recall that the tangents are constructed as lines perpendicular to the radii OQ and OR.) Conversely the image of P′ is P. In order to construct P given P′, we first join the line OP′. Construct the tangents from P′ to C (see Lemma 3 for this construction). The line QR intersects OP′ at the required point P.

The latter construction yields an algebraic formula for inversion: Note that the triangles OPQ and OQP′ are similar, since they share a common angle at O, and the corresponding angles at P and Q (respectively) are right angles. Therefore corresponding sides of the two triangles are in the same proportion, so that

Since OQ = r is just the radius of C, we obtain

Given that P′ lies on the line OP, the position of P′ is uniquely determined by its distance from O as given by this formula . We now prove a remarkable fact about pairs of inverse points:

**Theorem 1.** Let C be a circle, and let P and P′ be an inverse pair of
points with respect to CThen every circle through P and P′ is orthogonal to C.

In the special case that C has infinite radius (i.e. C is a Euclidean line) then inversion is simply reflection in this line, and the points P and P′ are mirror images in C.

Proof of Theorem 1. Consider any circle C′ through P and P′, and let T be the
center of C′. Let S be a point of intersection of C′ with T. In order to show
that the circles C and C′ are orthogonal, we only need to show that OS is
tangent to C′. Since the points P and P′ are inverse in C, we have OP · OP′ =
OS^{2}. Therefore
which, by Lemma 2, is exactly the length of the tangent from O to C′. Therefore
OS is

tangent to C′ as required.

**Theorem 2.** Inversion takes circles to circles.

Proof. Let C be a circle with center O. Let A and A′ be points inverted by C,
and let T and T′ be another pair of inverse points with respect to C. Let l be
the line passing through O, T and T′. Let m be the line passing through O, A and
A′. Let C_{1} be the unique circle through A tangent to l at T, and let C_{2} be the
unique circle through A′ tangent to l at T′ as shown. Let B be the second point
of intersection of m with C_{1}, and let B′ be the second point of intersection of
m with C_{2} . We must show that B′ is the inverse of B with respect to C, i.e. OB
· OB′ = r^{2} where r is the radius of C.

By Lemma 2 we have OT^{2} = OA · OB and (OT′)^{2} = OA′ · OB′.
Since T and T′ are inverse in C we have OT · OT′ = r^{2}, and similarly OA · OA′ =
r^{2}. Therefore

which yields r^{2} = OB · OB′ as required.

**Theorem 3.** Inversion preserves angles.

Proof. Consider a pair of points P and P′ inverted by a
circle C, and let C_{1} and C_{2} be two circles through P and P′. Let α and α′ be the
angles between C_{1} and C_{2}, at P and at P′ respectively, as shown. (This really
means the angles between the tangent lines to the circles C_{1} and C_{2}, at P and P′
respectively.) By Theorem 2, the inverse of C_{1} in C is a circle through P and
P′, meeting C at the same points as C_{1} does. But these points uniquely determine
the circle C_{1} , so the inverse of C_{1} with respect to C is C_{1} . Similarly the
inverse of C_{2} with respect to C is C_{2} . Therefore inversion takes angle α to
angle α′. However these two angles must be the same size by symmetry, since they
ar

e the angles between circles C_{1} and C_{2} at their two points of intersection. Thus
α′ = α as required.

Note that inversion, like reflection , reverses orientation of plane figures. The accompanying figure shows a letter ‘R’ and its inverse image in the circle C; note that in addition to distances being distorted, the orientation of the ‘R’ has been reversed.

**Interpreting Inversion in the Hyperbolic Plane**

Let C and C′ be a pair of orthogonal circles. Recall that the points of the plane interior to C represent points of the hyperbolic plane; and the arc of C′ interior to C represents a line of the hyperbolic plane. Inversion in C′ represents a reflection in the hyperbolic plane.

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