3 Orthogonal circles

Theorem 3.1. If two circles meet at P and Q, then the magnitude of the angles betweenthe circles is the same at P and Q.

Proof. Referring to the figure on the right,we have 4APB ≡ 4AQB (by SSS), so∠APB ≡ ∠AQB. Since the tangents to thecircles at P are perpendicular to the radiiAP and BP , it follows that the angle be-tween the tangents at P is equal in mea-sure to m(∠APB). Likewise, the angle be-tween the tangents at Q is equal in measureto m(∠AQB).

A B

P

Q

Remark: The previous theorem means that to check the angle between two circles in-tersecting at P and Q, we only need to check one of the angles. Note however, that thedirections of the angles at P and Q are opposite.

Definition: Two intersecting circles α and β are said to be orthogonal if the angle betweenthem is 90◦. We sometimes write α ⊥ β to indicate orthogonality.

Theorem 3.2. If two circles α and β are orthogonal, then

(1) The tangents at each point of intersection pass through the centres of the other circle.(Figure (a) below).

(2) Each circle is its own inverse with respect to the other.

O

T

PQ

!"

!"

(a) (b)

Proof.

(1) This follows because a line through the point of tangency perpendicular to the tangentmust pass through the centre of the circle.

(2) Let P be a point on the circle β. Join P to O, the centre of α, and let r be the radiusof α. Let Q be the other point where the ray OP meets β. Let T be the point ofintersection of the two circles so that OT is the tangent to β by part (1). By the powerof the point O with respect to β we have

OP ·OQ = OT 2 = r2 ,

showing that the inverse of any point on β is another point on β.

9

The preceding theorem has the following converse:

Theorem 3.3. Two intersecting circles α and β are orthogonal if any one of the followingstatements is true.

(1) The tangents to one circle at one point of intersection passes through the centre of theother circle (Figure (a) below).

(2) One of the circles passes through two distinct points that are inverses with respect tothe other circle.

O

X

PQ

!"

!"

(a) (b)

r

Proof.

(1) This implies that the two tangents at the point of intersection must be perpendicular.

(2) Suppose that circle β passes through P and Q that are inverses with respect to α. LetO be the centre of α and let OX be tangent to β at X (see Figure (b)). Then

OP ·OQ = OX2 (by the power of O with respect to β)

OP ·OQ = r2 (since P and Q are inverses).

This implies that OX = r, so X must be on α as well as on β. That is, X is a point ofintersection of α and β, and the tangent to β at this point passes through the centreof α. By part (1), the circles must be orthogonal.

10

The Arbelos Theorem (Ogilvy, p. 54; Eves, p. 133)

Theorem 3.4. (The Arbelos Theorem a.k.a. Pappus’ Ancient Theorem). P , Q, and R arethree collinear points with C, D, and K0 being semicircles on PQ. PR, and QR. Let K1,K2, . . ., be circles touching C and D with K1 touching K0, K2 touching K1 and so on.

Let hn be the distance of the centre of Kn from PR and let rn be the radius of Kn. Thenhn = 2nrn.

K0

K1

K2Kn

hn

C

D

P Q R

Proof.

K0

K1

K2

Kn

C

D

l m

P Q R

Let t be the tangential distance from P to the circle Kn, and apply I(P, t2).

Kn is orthogonal to the circle of inversion, so is its own inverse.

C inverts into a line l.

D inverts into a line m parallel to l.

K0 inverts into a semicircle K ′0 tangent to l and m (because inversion preserves tangencies).

Ki inverts into a circle K ′i tangent to l and m.

Then all of the K ′i have the same radius, namely rn, and the theorem follows.

11

Steiner’s Porism (Ogilvy, p. 51–54; Eves, p. 134, 135)

Given a point P outside a circle α, a point X of α is said to be visible from P if the segmentPX meets α only at X.

!

X

X

P

not visible from P

visible from P !

P

(a) (b )

Figure (b) above shows in red the set of points that are visible from P , namely the twotangent points and the points on the arc between the tangent points. In other words, apoint X of α is visible from P if and only if either

PX is tangent to α, or

the tangent to α at X has α and P on opposite sides.

Note also that if a line m is tangent to α at X, and if P is on the same side of m as α (butnot on the line m), then X is not visible from P .

Lemma 3.5. Suppose the line PQ misses the circle α. Then

(1) There is a point X visible from both P and Q.

(2) There is a point Y visible from P but not from Q.

(3) There is a point Z visible from Q but not from P .

The figure below illustrates how to find points X and Y : Let m be a line parallel to PQ andtangent to α. X is the point of tangency of m with α. There are two lines from P tangentto α. Let l be the tangent line such that α and Q are both on the same side of l. Then Yis the point where l is tangent to α.

X

P Q

m

Y

!l

12

Lemma 3.6. Given two circles ω and ξ, and given a point P not on either circle, then thereis a circle through P orthogonal to both ω and ξ.

Proof.

! !

""

PP

QQ

S SO OU U

Let Q be the inverse of P with respect to ω. Let S be the inverse of P with respect to ξ.Then there is a unique circle through P , Q, and S, and this circle must be orthogonal toboth ω and ξ by Theorem 3.3.

Note: If O, U , and P are collinear, then the orthogonal “circle” is a line. If O, U , and Pare not collinear, then the orthogonal circle is an true circle.

Lemma 3.7. Let ω and ξ be two non-intersecting circles with centres O and U , O 6= U .Then we can find points X and Y that are inverses to each other with respect to both ωand ξ.

Proof. Let α be any circle other than a line that is orthogonal to both ω and ξ. We claimthat the line OU intersects α in two points. (Then the two points are X and Y and theyare inverses to each other with respect to both ω and ξ.)

! !

""

X YY XO OU U

case (i) case (ii)

##

There are two cases to consider: (i) when the circles are exterior to each other and (ii) whenone circle is inside the other.

(i) Suppose, for a contradiction, that OU misses α. Then there is a point Z on α thatis visible from both O and U . Since Z is visible from O, it is inside or on ω. SinceZ is visible from U , it is inside or on ξ. Then Z is inside or on both ω and ξ, whichcontradicts the fact that ω and ξ are exterior to each other. This proves case (i).

(ii) Left as an exercise.

13

Definition. Let ω and ξ be two non-intersecting circles.Let α1 be a circle tangent to both ω and ξ.Let α2 be a circle tangent to α1, ω, and ξ.Let α3 be a circle tangent to α2, ω, and ξ.Continue in this fashion. If at some point αk is tangent to α1 we say that α1, α2, . . . , αk

is a Steiner chain of k circles.

!

"

Remark: Given two circles ω and ξ, there is no guarantee that a Steiner chain exists for ωand ξ.

Theorem 3.8. (Steiner’s Porism.) Suppose that the two non-intersecting circles ω and ξhave a Steiner chain of k circles. Then, any circle tangent to ω and ξ is a member of someSteiner chain of k circles.

To prove this theorem, we need the following:

Lemma 3.9. Given two non-intersecting circles ω and ξ (that are not concentric), there isan inversion that transforms them into concentric circles.

Proof. Using Lemma 3.6 we can find two circles α and β simultaneously orthogonal to bothω and ξ. These two circles intersect at points X and Y referred to in Lemma 3.7.

!

"

X YO U

#

$

'#

'" '!

'$

Y'

Perform the inversion I(X, r2) for some radius r. Thenα transforms to α′ (a straight line through Y ′ and not through X).β transforms to β′ (a straight line through Y ′ and not through X).ω transforms to a circle ω′, and ω′ ⊥ α′, ω′ ⊥ β′(because orthogonality is preserved).ξ transforms to a circle ξ′, and ξ′ ⊥ α′, ξ′ ⊥ β′.

Since the circle ω′ is orthogonal to the line α′, then ω′ must be centred at some point of α′.Similarly, ω′ must be centred at some point of β′. Thus, ω′ is centred at Y ′. By the sameargument, ξ′ must also be centred at Y ′.

14

Proof of Steiner’s Porism.

!e circle is obviously part of a Steiner chain of k circles, so by the reverse inversion mustalso be part of a Steiner chain of k circles.

!

Invert and into concentric circles. !e inversion preservesthe Steiner chain of k circles.

! ""

'!

'"

!

"

'!

'"

'#

'#

Using the same inversion,transform into a circle .'##

#

#

15