MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 ·...

54
MATH 3210: Euclidean and Non-Euclidean Geometry Circular Inversion Images of Lines and Circles April 13, 2020 Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Transcript of MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 ·...

Page 1: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

MATH 3210:Euclidean and Non-Euclidean Geometry

Circular Inversion

Images of Lines and Circles

April 13, 2020

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 2: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 3: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 4: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F .

Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

O

AA′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 5: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O

the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 6: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 7: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 8: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.

• If AρΓ7→ A′, then A′ ρΓ7→ A.

• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 9: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.

• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 10: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Circular Inversion

We will work in a Cartesian plane ΠF over a Euclidean ordered field (F ;<).

Recall: ΠF is s Euclidean plane ( = a Hilbert plane satisfying axioms (P) and (E)).

Definition. Let Γ be a circle with center Oand radius of length r ∈ F . Circular inver-sion with respect to Γ is the mappingρΓ : ΠF \ {O} → ΠF \ {O} that assignsto every point A 6= O the unique point A′

on ray−→OA such thatdist(O,A) · dist(O,A′) = r2,

which we will abbreviate as:OA ·OA′ = r2.

Γ

OA

A′

• Every point of Γ is fixed by ρΓ.• If A

ρΓ7→ A′, then A′ ρΓ7→ A.• ρΓ switches the inside of Γ (with O removed) and the outside of Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 11: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P.

O

Γ

A A′

P

QProof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 12: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ.

Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P.

O

Γ

A A′

P

QProof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 13: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P.

The following are equivalent:(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P.

O

Γ

A A′

P

QProof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 14: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),

(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P.

O

Γ

A A′

P

QProof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 15: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

QProof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 16: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).

(2)⇒ (1): (2)⇒ OP2

= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 17: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 18: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 19: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:

� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 20: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;

� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 21: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 22: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:

� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 23: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

Q

Proof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;

� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 24: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Constructing the Image of a Point by Ruler and Compass

Let Γ be a circle with center O and radius of length r ∈ F .

Theorem 1. Let A, A′ be points such that−→OA =

−−→OA′, A is inside Γ,

and A′ is outside Γ. Let the line through A, perpendicularto OA, meet Γ at P. The following are equivalent:

(1) A′ = ρΓ(A),(2) ∠OPA′ is a RA,(3) line A′P is tangent to Γ at P. O

Γ

A A′

P

QProof. (2)⇔ (3) by (III.16), (III.18).(2)⇒ (1): (2)⇒ OP

2= OA · OA′ by the proof of (I.47) (in last lecture)⇒ (1).

(1)⇒ (2): (1)⇒ OPOA′ = OA

OP

(Sim SAS)⇒ 4OAP ∼= 4OPA′ ⇒ RA ∠OAP ∼= ∠OPA′ ⇒ (2).

Construction of A′ from A:� construct perpendicular line through A to line OA, get P;� construct perpendicular line through P to line OP, get A′.

Construction of A from A′:� construct circle with diameter OA′, get P,Q;� draw line PQ, get A.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 25: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O.

O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 26: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O.

O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 27: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O.

O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 28: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,

(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O.

O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 29: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,

(3) circle containing O ρΓ7→ line not containing O.

O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 30: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 31: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.

(2)–(3) Given a line ` s.t. O /∈ `,– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 32: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 33: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

A

A′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and

– let γ be the circle with diameter OA′ where A′ = ρΓ(A).Vice versa, given a circle γ s.t. O ∈ γ,

– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 34: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 35: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,

– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 36: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and

– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 37: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 38: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 39: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,

• OAOB′ = OB

OA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 40: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Lines and Images of Circles Through O

Let Γ be a circle with center O and radius of length r ∈ F .

Note: O is removed from circles and lines when we apply ρΓ, since ρΓ(O) is undefined.

Theorem 2.(1) line containing O ρΓ7→ itself,(2) line not containing O ρΓ7→ circle containing O,(3) circle containing O ρΓ7→ line not containing O. O

Γ

`

AA′

γ

X

B′

B

Proof.(1) Clear from the definition of ρΓ.(2)–(3) Given a line ` s.t. O /∈ `,

– let A ∈ ` be the foot of the perpendicular to ` through O, and– let γ be the circle with diameter OA′ where A′ = ρΓ(A).

Vice versa, given a circle γ s.t. O ∈ γ,– let OA′ be the diameter of γ through O, and– let ` be the line through A = ρΓ(A′) perpendicular to line OA′.

In either case:• the rays

−→OB (B ∈ `), via meeting with γ, set up a bijection `→ γ \ {O}, B 7→ B′.

• 4OAB ∼ 4OB′A′ by (Sim AAA), using (III.31) and (I.32). Hence,• OA

OB′ = OBOA′ , i.e., r2 = OA · OA′ = OB · OB′, so B′ = ρΓ(B) and B = ρΓ(B′).

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 41: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 42: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 43: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 44: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:

(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C

γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 45: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;

(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C

γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 46: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);

(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C

γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 47: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C

γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 48: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

P

Proof. (b)⇒ (c) is clear. Let C denote the center of γ.

(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 49: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 50: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 51: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 52: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 53: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Images of Circles Perpendicular to Γ

As before, let Γ be a circle with center O and radius of length r ∈ F .

Definition. If two circles [a line and a circle] meet at a point P, theangle between them at P is defined as the angle between theirtangent lines at P [the line and the tangent line of the circle].

Note. The angles at the two points of intersection are congruent.

Theorem 3. For any circle γ, the following conditions are equivalent:(a) γ is perpendicular to Γ;(b) ρΓ maps γ onto itself (i.e., ρΓ(γ) = γ);(c) γ contains two distinct points A,A′ such that ρΓ(A) = A′.

O

Γ

C γ

A

A′

X

PProof. (b)⇒ (c) is clear. Let C denote the center of γ.(a)⇒ (b):

γ ⊥ Γ(III.16,18)⇒ γ meets Γ at a point P such that OP ⊥ CP

(III.16,36)⇒ OA · OA′ = r2 for any ray−→OX meeting γ at A,A′.

(c)⇒ (a):ρΓ(A) = A′ ⇒ one of A,A′ is inside, the other is outside Γ

(E)⇒ γ meets Γ at a point P and OA · OA′ = r2

conv. of (III.36)⇒ OP is tangent to γ at P(III.16,18)⇒ γ ⊥ Γ.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry

Page 54: MATH 3210: Euclidean and Non-Euclidean Geometryszendrei/Geom_S20/lec-04-13.pdf · 2020-04-13 · Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry. Constructing the

Practice Problems on Circular Inversion

We work in the Cartesian plane ΠF over a Euclidean ordered field F .Let Γ be a circle with center O and radius of length r ∈ F .

Problem 1. Which lines are perpendicular to Γ?

Problem 2. Suppose that a line ` meets Γ at two distinct points A,Bsuch that O is not on `. Show that ρΓ(`) is the (unique) circle thatpasses through A, B, and O.

Circular Inversion MATH 3210: Euclidean and Non-Euclidean Geometry