CUESTIONARIO BASE PARA DEFINICIÓN DEL ALCANCE

Assume two circlesΓand Ωintersect at two points X andY. Letℓ and

mbe the tangent lines atX toΓandΩcorrespondingly. Analogously,ℓ′ andm′ _{be the tangent lines atY toΓandΩ.}

We say that the circleΓisperpendicular to the circleΩ(brieflyΓ⊥Ω) if they intersect and the lines tangent to the circles at one point (and therefore, both points) of intersection are perpendicular.

Similarly, we say that the circleΓis perpendicular to the lineℓ(briefly Γ ⊥ ℓ) if Γ∩ℓ 6= ∅ and ℓ perpendicular to the tangent lines to Γ at one point (and therefore, both points) of intersection. According to Lemma 5.16, it happens only if the lineℓpasses thru the center ofΓ.

Now we can talk aboutperpendicular circlines.

10.15. Theorem. AssumeΓ andΩare distinct circles. ThenΩ⊥Γ if and only if the circle Γ coincides with its inversion inΩ.

O Q

X Y

Proof. LetΓ′ _{denotes the inverse of Γ.}

“Only if ” part. LetO be the center of ΩandQ be the center of Γ. Let X andY denote the points of intersections ofΓandΩ. According to Lemma 5.16, Γ ⊥ Ω if and only if (OX) and (OY) are tangent toΓ.

Note thatΓ′ _{is also tangent to(OX)and (OY)}

atX andY correspondingly. It follows thatX andY are the foot points of the center of Γ′ _{on (OX)and (OY). Therefore, both Γ}′ _{and Γ have} the centerQ. Finally, Γ′_{= Γ, since both circles pass thruX.}

“If ” part. Assume Γ = Γ′_.

SinceΓ6= Ω, there is a pointPthat lies onΓ, but not onΩ. LetP′_be the inverse ofP in Ω. SinceΓ = Γ′_{, we have thatP}′ _{∈Γ. In particular,} the half-line [OP) intersects Γ at two points. By Exercise 5.12, O lies outside ofΓ.

AsΓhas points inside and outside ofΩ, the circlesΓandΩintersect. The latter follows from Exercise 3.20.

LetX be a point of their intersection. We need to show that(OX)is tangent toΓ; that is,X is the only intersection point of(OX)andΓ.

AssumeZ is another point of intersection. SinceO is outside ofΓ, the pointZ lies on the half-line[OX).

LetZ′_{denotes the inverse ofZinΩ. Clearly, the three pointsZ, Z}′_{, X} lie onΓand(OX). The latter contradicts Lemma 5.14.

It is convenient to define theinversion in the line ℓas the reflection acrossℓ. This way we can talk aboutinversion in an arbitrary circline. 10.16. Corollary. Let Ω and Γ be distinct circlines in the inversive plane. Then the inversion inΩsends Γto itself if and only if Ω⊥Γ. Proof. By Theorem 10.15, it is sufficient to consider the case when Ωor Γ is a line.

Assume Ωis a line, so the inversion in Ω is a reflection. In this case the statement follows from Corollary 5.7.

IfΓis a line, then the statement follows from Theorem 10.11. 10.17. Corollary. Let P and P′ _{be two distinct points such that P}′ _is the inverse of P in the circle Ω. Assume that the circline Γ passes thru

P andP′_{. ThenΓ⊥Ω.}

Proof. Without loss of generality, we may assume thatP is inside andP′ is outsideΩ. By Theorem 3.17,Γ intersectsΩ. LetAdenotes a point of intersection.

Let Γ′ _{denotes the inverse of Γ. Since A is a self-inverse, the points}

A,P, andP′ _{lie onΓ}′_{. By Exercise 8.2,Γ}′ _{= Γand by Theorem 10.15,} Γ⊥Ω.

10.18. Corollary. Let P andQbe two distinct points inside the circle Ω. Then there is a unique circline Γ perpendicular toΩ that passes thru

P andQ.

Proof. LetP′ be the inverse of the pointP in the circleΩ. According to Corollary 10.17, the circline is passing thruP andQis perpendicular to Ωif and only if it passes thru P′_.

Note thatP′ _{lies outside ofΩ. Therefore, the pointsP,P}′_{, andQare} distinct.

According to Exercise 8.2, there is a unique circline passing thru P,

Q, and P′_{. Hence the result.}

10.19. Exercise. LetP,Q,P′_{, andQ}′ _{be points in the Euclidean plane.} AssumeP′ _andQ′ _{are inverses of P and Q correspondingly. Show that} the quadrilateralP QP′Q′ is inscribed.

10.20. Exercise. Let Ω1 andΩ2 be two perpendicular circles with cen-

ters at O1 and O2 correspondingly. Show that the inverse of O1 in Ω2

coincides with the inverse ofO2 in Ω1.

10.21. Exercise. Three distinct circles —Ω1,Ω2, and Ω3, intersect at

two points —AandB. Assume that a circleΓis perpendicular toΩ1and

Ω2. Show thatΓ⊥Ω3.

Let us consider two new construction tools: thecircum-tool that con- structs a circline thru three given points, and theinversion-tool — a tool that constructs an inverse of a given point in a given circline.

10.22. Exercise. Given two circlesΩ1,Ω2 and a pointP that does not

lie on the circles, use only circum-tool and inversion-tool to construct a circline Γthat passes thruP, and perpendicular to bothΩ1 andΩ2.

10.23. Advanced exercise. Given three disjoint circles Ω1, Ω2 and

Ω3, use only circum-tool and inversion-tool to construct a circline Γthat

perpendicular to each circle Ω1,Ω2, andΩ3.

Think what to do if two of the circles intersect.