circle lab

8/12/2019 Circle Lab

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98 CHAPTER 2. EUCLIDEAN GEOMETRY

Exercise 2.6.13. Show the converse to the preceding exercise, that is, that the

bisector of the angle made by two tangents from a point outside a circle to the circlemust pass through the center of the circle. [Hint: Try a proof by contradiction.]

Exercise 2.6.14. Let c and c be externally tangent at T. Show that there aretwo lines that are tangent to both circles (at points other than T). [Hint: Letmbe the line through the centers. Consider the two radii that are perpendicular tom. Let l be the line through the endpoints of these radii on their respective circles.If l and m are parallel, show that l is a common line of tangency for both circles.If l and m intersect at P, let n be a tangent from P to one of the circles. Show nis tangent to the other circle.]

2.7 Project 4 - Circle Inversion and Orthogonality

In this project we will explore the idea of inversion through circles. Circleinversion will be a critical component of our construction of non-Euclideangeometry in Chapter 7.

We start out with the notion of the powerof a point with respect to agiven circle.

Start the Geometry Explorer pro-gram and create a circle c with cen-ter Oand radius point A, and createa point Pnot on c.

O

A

c

P


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2.7. PROJECT 4 - CIRCLE INVERSION 99

Now create two lines originating

at P that pass through the circle.Find the two intersection points ofthe first line with the circle (callthem P1 and P2) and the two in-tersection points of the second linewith the circle (call them Q1 andQ2). Measure the four distancesP P1, P P2, P Q1, and P Q2. (Tomeasure distance, multi-select twopoints and chooseDistance(Mea-suremenu).)

O

A

c

P

P2

P1

Q2

Q1

Dist(P,P1) = 1.83

Dist(P,P2) = 6.00Dist(P,Q1) = 1.82

Dist(P,Q2) = 6.01

Now we will compare the prod-uct ofP P1 and P P2 to the productof P Q1 and P Q2. To do this wewill use the Calculator in GeometryExplorer. Go to the Help Web page(click onHelpin the menu bar) andthen go to the View Menu linkand from there to the Calculatorlink. Read through this section to

become familiarized with how to usethe Calculator. Now, choose Cal-culator(View menu).

Notice that the four distance measurements are listed in the right half ofthe Calculator window. Double-click the first distance measure, then clickon the Multiplication button (labeled *), and then double-click the seconddistance measurement. We have just created an expression for the productofP P1 and P P2.


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To use this measurement back in

the Geometry Explorer main win-dow, we click the Evaluate buttonand then the Add to Canvas but-ton. The new product measure willnow be on the screen. Do the samefor the product of P Q1 and P Q2.[Be sure to Clear the Calculatorfirst.]

O

A

c

P

P2

P1

Q2

Q1

Dist(P,P1) = 1.83

Dist(P,P2) = 6.00

Dist(P,Q1) = 1.82

Dist(P,Q2) = 6.01

Dist(P,P1) *Dist(P,P2) = 10.95

Dist(P,Q1) *Dist(P,Q2) = 10.95

Interesting! It appears that these two products are the same. Drag pointParound and see if this conjecture is supported.

Exercise 2.7.1. Our first task in this project is to provethat these two productsare always the same. [Hint: Consider some of the inscribed angles formed byP1, P2, Q1,Q2. Use Corollary 2.32 to show that PP1Q2 is similar to PQ1P2and thus show the result.]

Exercise 2.7.2. Show that the product ofP P1 and P P2 (or P Q1 andP Q2) canbe expressed as PO2 r2, where r is the radius of the circle.

Definition 2.34. Given a circle c with center O and radius r and given apoint P, we define the Power ofPwith respect to c as:

P ower of P=P O2 r2.

Note that by Exercise 2.7.2 the Power ofP is also equal to the productofP P1 and P P2 for any line l from P, with P1 andP2 the intersections oflwith the circle c.

Also note that the Power ofP can be used to classify whetherP is inside(Power 0) the circle.

Now we are ready to define circle inversion.

Definition 2.35. Theinverse ofPwith respect to c is the unique point P

on ray

OP such that OP = r2

OP (or (OP)(OP) =r2).

Note that if the circle had unit radius (r = 1), and if we considered Oas the origin in Cartesian coordinates with OP =x, then the inverse P ofPcan be interpreted as the usual multiplicative inverse; that is, we wouldhave OP = 1

x.

How do we construct the inverse point?


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Clear the screen and create a circle

c with center O and radius point Aand then create a point P inside c.

Create the ray

OP. AtP construct

the perpendicular to

OP and findthe intersection points (T and U)of this perpendicular with the cir-cle. Create segment OT and findthe perpendicular to OT at T. LetP be the point where this second

perpendicular intersects

OP.

O

A

c

P

T

U

P

Measure the distances for seg-ments OP and OP and measurethe radius of the circle. Use the Cal-culator to compute the product ofOP and OP and the square of theradius as shown in the figure.

O

A

c

P

T

U

P

Dist(O,P) = 1.24Dist(O,P) = 5.10

Radius(c) = 2.52

Dist(O,P) *Dist(O,P) = 6.33Radius(c) ^2 = 6.33

It appears that we have constructed the inverse!

Exercise 2.7.3. Prove that this construction actually gives the inverse of P.That is, show that (OP)(OP) = r2.

In the last part of this lab, we will use the notion of circle inversion toconstruct a circle that meets a given circle at right angles.

Definition 2.36. Two circles c and c that intersect at distinct points Aand B are called orthogonal if the tangents to the circles at each of these

points are perpendicular.

Suppose we have a circle c and two points P andQ insidec, with P notequal to Q and neither point equal to the center O of the circle. The goalis to construct a circle through P and Q that meets c at right angles.


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Using the ideas covered earlier in this project, construct the inverse P

of P with respect to c. Then, select P, P

, and Q and click on the Circletool in the Construct panel to construct the unique circle c through thesethree points. The claim is that c is orthogonal to c.

To see if this is the case, lets first find the center of c. Let R be the

intersection of

T Pwith circle c (Fig 2.32).

O

c

P

T

P

Q

c

S1

S2

R

O

Angle(O,S1,O) = 90.00 degrees

Angle(O,S2,O) = 90.00 degrees

Fig. 2.32

Then RP P is a right angle in circle c as OP T is a right angle.Thus, by Theorem 2.33 RP is a diameter of c. The midpoint O of RP

will be the center ofc. Let S1 and S2 be the intersection points ofc withc. Measure OS1O and OS2O

and check that they are right angles.Since the tangents to c and c are orthogonal to OS1, OS2, O S1, and O S2,then the tangents to the circles at S1 and S2 must also be orthogonal andthe circles are orthogonal. Note that this evidenceof the orthogonality ofc and c is not a rigorous proof. The proof will be covered when we get toTheorem 2.38.

Exercise 2.7.4. What do you think will happen to circle c as one of the pointsP orQ approaches the center O of circle c? Try this out and then explain why thishappens.


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Project Report

The ability to construct orthogonal pairs of circles is crucial to developing amodel of hyperbolic geometry, where parallels to a line through a point areabundant. We will look at this model in detail in Chapter 7.

For the project report, provide detailed analysis of the constructionsused in this project and complete answers to the exercises.

2.7.1 Orthogonal Circles Redux

Here is a proof of orthogonality of the circles constructed in the text pre-

ceding Exercise 2.7.4.

Theorem 2.38. Given a circle c with center O and radius OA and giventwo pointsP andQ insidec, withPnot equal to Q and neither point equalto O, there exists a unique circle c (or line) that passes through P and Qthat is orthogonal to the given circle (Fig. 2.33).

O

A

c

PP

Q

c

T

l

Fig. 2.33

Proof: It is clear that if P and Q lie on a diameter of c, then there isa unique line (coincident with the diameter) that is orthogonal to c. So, inthe rest of this proof we assume that P and Q are not on a diameter ofc.

Suppose that one or the other ofP or Q, say P, is strictly inside c. Asabove construct the inverse P to P and let c be the unique circle passing


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through Q,P,P. Construct a tangent l to circle c that passes through O.

Let l be tangent to c

at T. (To construct l, use the construction discussedin Theorem 2.37.) We claim that T is also on circle c. To see this, considerthe power ofO with respect to circle c:

P ower of O = (OP)(OP) = (OT)2

But, (OP)(OP) = r2 (r being the radius of c) since P is the inversepoint to P with respect to c. Thus, (OT)2 = r2 and T is on circle c, andthe circles are orthogonal at T.

To see that this circle is unique, suppose there was another circle c

through P and Q that was orthogonal to c. Let P be the intersection of

OPwith c. Let T be a point where c and c intersect. Then (OP)(OP) =

(OT

)2

. But, (OT

)2

= r2

and thus, P

must be the inverse P

to P, andc must then pass through Q, P,P and must be the circle c.

The final case to consider is when both P and Q are on the boundary ofc(Fig. 2.34). Then, any circle throughP andQ that is orthogonal toc musthave its tangents at P and Q lying along OP and OQ. Thus, the diametersof this circle must lie along tangent lines to c at P and Q. Thus, the centerof the orthogonal circle must lie at the intersection of these tangents, whichis a unique point.

O

c

P

Q

O

c

Fig. 2.34

We conclude this section on orthogonal circles with two results that willprove useful when we study non-Euclidean geometry in Chapter 7.


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Theorem 2.39. Letc andc be two circles and letPbe a point that is not

onc and is not the centerO ofc. Suppose thatc

passes throughP. Then,the two circles are orthogonal if and only if c passes through the inversepointP to Pwith respect to c.

Proof: First, suppose thatc passes through the inverse point P (referto Fig. 2.35). We know from the proof of Theorem 2.30 that the centerO

ofc lies on the perpendicular bisector ofP P. Since P and P are inverses

with respect to c, then they both lie on the same side of ray

OP. Thus, Ois not between P and P and we have that OO > OP. Thus, O is outsideof c. We then can construct two tangents from O to c at points T1 andT2 on c

. Using the idea of the power of points with respect to c, we have(OT1)

2 = (OP)(OP). But, (OP)(OP) =r2 by assumption, where r is the

radius of c. Therefore, (OT1)2 = r2, and T1 is on c. A similar argumentshows that T2 is also on c. This implies that the two circles are orthogonal.

O

cT1

T2

O

c

P

P

Fig. 2.35

Conversely, suppose that c and c are orthogonal at points T1 and T2.The tangent lines to c at these points then pass through O, which implies

that O is outside c. Thus,

OPmust intersect c at another point P. Usingthe power of points, we have r2 = (OT1)

2 = (OP)(OP), and thus P is theinverse point to Pwith respect to circle c.

Corollary 2.40. Suppose circles c and c intersect. Then c is orthogonalto c if and only if the circlec is mapped to itself by inversion in the circlec.

Proof: Suppose the circles are orthogonal, and let Pbe a point on c . IfP is also on c, then it is fixed by inversion through c. IfP is not on c, then


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by the proof of Theorem 2.39, we know that Pis also not the centerO ofc.

Thus, Theorem 2.39 implies that the inverseP

ofPwith respect to c is onc. Thus, for all points P on c , we have that the inverse point to P is againon c.

Conversely, supposec is mapped to itself by inversion in the circle c. LetPbe a point onc that is not onc and which is not the centerO ofc. Then,the inverse point P with respect to c is again on c. By Theorem 2.39, thecircles are orthogonal.

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