Notes — Chapter 12: Circles
To the Student. The sun or moon appear as disks. A drop of oil on water spreads out in a circle. Wheels are round. A circle is defined by its center and its radius. A circle has perfect symmetry about its center and is sent onto itself by reflection in any line through its center and by rotations about its center. Your blood pushes outward in all directions equally, so arterial cross-sections are circular. In order for the bearing on an axel to allow the wheel to turn freely, it must have rotational symmetry, so it must be circular.
Here we study this simplest curved shape, its uses, and its properties.
Lesson 1. Circles, Radii, and Chords. Here we formally meet the circle and its radii, diameters, and chords. We learn that if a line through the center of a circle is perpendicular to a chord, it bisects that chord (Theorem 56) and that if a line through the center bisects a chord that is not a diameter, it is perpendicular to the chord (Theorem 57). These results are combined in Proposition 3 of Book III of Euclid.
Set I exercises cover the definitions and key ideas. Exercises 16-26 give direct proofs of Theorems 56 and 57 and are good as spiral review and for learning more about circles.
Exercises 27-29. The indirect proof Theorem 58, which states that the perpendicular bisector of a chord goes through the center of the circle, can be replaced by the following direct proof. If AB is the chord of circle O, then the center O is equidistant from A and B, as is C, the point that bisects AB. Hence, by Theorem 16, page 212, O and C determine the perpendicular bisector of AB. Hence O lies on the perpendicular bisector of the chord AB, as was to be shown.
Set ITI has many applications, including the argument Ole Romer (1644-1710) used to make the first determination of the speed of light, Exercises 37-39. The Three Circles figure is quite surprising when completed, Exercises 43-47.
Set III. Island Puzzle yields to a nice “Aha!”
Lesson 2. Tangents. We open this lesson with the hammer throw. The proof that a tangent is perpendicular to the radius to the point of tangency (Theorem 59) is direct and geometric. A proof that a line ¢ perpendicular to a radius OA at A (Theorem 60), can be given using the Pythagorean Theorem. If B # A is on é then OB = OA? + AB? > OA. So B lies strictly outside of circle O, and & touches the circle O only at A. Thus @ is tangent to O at A.
Algebraic methods based on established results are as valid as geometric ones. The choice of method of proof
depends on personal preference or on which sort of reasoning seems easiest to follow for a given result. What is clearest for me may not be clearest for you.
Set I. This covers the core material for the lesson. Tangents allow us to smoothly piece together portions of circles and segments. See Band Saw (1-3) and Railroad Curve (19-23). In practice, railroad curves are designed to be even smoother than those described in (19-23). They keep the “jerk derivative” within bounds. This is the derivative of the acceleration. When a railroad car rounds a curve, the radius of curvature, r, of the track and velocity, v, of the train determine the “centrifugal force,” which has magnitude v?/r. This is the force that makes you lean toward the outside of a curve on a train. If this force shifts suddenly, the passengers will lurch from side to side in the car. That would not be good. So engineers need to make the lead-in to a curve more gentle than our considerations suggest. You can look up jerk derivative on the Web if you want more information about this.
Set III asks about a point from which the earth and moon have the same apparent diameter. You might ask your students if they think there are other points with this property.
There is a second such point beyond the moon, where the moon just eclipses the earth. The locus of all such points is a circle of Apollonius, a circle on which the ratio of distances to the centers of the moon and of the earth remains constant. If d, is the distance from our point P to the center of the earth and d,, the distance to center of the moon, while r, and 7r,, are the radii of the earth and moon, the condition de/dm = Tm/Te will insure that the two have the same apparent diameter. Inversion in this circle of Apollonius interchanges the points that are the centers of the earth and moon. Search on circles of Apollonius or inversive geometry at websites such as Eric Weisstein’s MathWorld. See also, Geometry, Hollow Earth, page 668 for the inversion in a sphere that interchanges points P and P’.
Lesson 3. Central Angles and Arcs. Arcs and arc measure are properly introduced here along with reflex angles. Intuition and common practice let us deal with clockwise or counterclockwise rotations through any number of degrees. For technical reasons, Euclid stuck to angles no larger than a straight angle. We need angles larger than 180° to deal with major arcs. Postulate 10, The Arc Addition Postulate, can be proven from the Protractor Postulate by analyzing the different possibilities for arrangements of minor and major arcs determined by points A, B, and C.
▶ Curvature and the Measure of Arcs. The local curvature at each point of a circular arc is 1/r, where r is its radius. The product of the length, S, of the arc by its curvature gives the total curvature of the arc, S/r. This total curvature is the radian measure of angle through which the arc’s tangent turns as the
arc is traversed. Using calculus, we can define the radius of curvature, r, for a smooth curve at each of its points and a corresponding curvature, 1/r, at each point of the curve, and the radian measure of the angle through which the tangent of the curve turns as we move along any stretch of the curve is the integral of ds/r, where ds is the element of length for the curve.
The tangent may turn counterclockwise or clockwise depending on the direction of travel along the arc and on which way the arc is bowed. For a closed curve, we can take this into account by taking the curvature to be = 1/r when the arc is bowed outward and −1/r when the arc is bowed inward. This will make our accounting come out right no matter which way we go around the curve.
For a smooth simple closed curve, the integral of ds/r around a single circuit must be 27 radians. This is because, taking clockwise and counterclockwise turnings with the proper accounting of signs, the tangent must turn through a total of one full rotation. To convert to degrees, multiply by 360°/27 in all calculations and the integral becomes 360°.
If we are dealing with arcs of circles smoothly put together, this integral becomes a sum over the arcs of the arc measures, taking the direction of turning (clockwise or counterclockwise) into account. This is why the sum of the arc measures in The Square That Isn’t (30-36) must be 360°. The same argument holds for the arc measures in the Set III exercise, so Obtuse Ollie gets this one right.
Set I. Nice coverage of key ideas. Arc Illusion (19-22) is provocative.
Set II. Note the original proofs, Exercises 43-47. The Night Sky (40-42) asks intriguing questions.
Lesson 4. Inscribed Angles. “In a circle the angle at the center is double the angle at the circumference, if the angles intercept the same arc” as we read in Euclid, Book III, Proposition 20. This becomes Theorem 63 when stated in terms of inscribed angles and arc measure. Corollary 1 to Theorem 63 tells us that inscribed angles that intercept the same arc are equal (Euclid Book III, Proposition 21). These are useful in applications and lead to further by allowing us to show that various triangles connected with circles are similar. See Exercises 55-58 of this lesson and Exercises 6-10 of Lesson 6, page 512.
The proof of Theorem 63 (6-19) uses algebra applied to angle measures.
Basic ideas are covered in Set I exercises. Set II, intersecting Triangles and Circles (55-58) is beautiful, and Set III’s Finding the Center is a practical problem that gives a nice “Aha!”
Lesson 5. Secant Angles. Given two lines that both cut a circle, we can find the angle between them
by knowing the arcs they cut off between them. The proofs of formulas are given in Exercises 1-12, and they offer practice with the ideas of the previous lesson.
Here is a more dynamic way to look at these theorems. Look at Figure 1 on page 511. Ask yourself: How much angular rotation is needed to turn line DC counterclockwise so as to coincide with line EA? The process is done in two stages. Keep D fixed as a pivot and rotate C around so the line runs through A. You have a counterclockwise rotation of 80°/2 = 40°, by Theorem 63. Now your line runs through D and A. Pivoting it about A, rotate the lower part counterclockwise of it until it passes through E. This is an additional counterclockwise rotation of 40°/2 = 20°. So the total counterclockwise rotation is 40°+20° = 60°. This is the rotation required to bring one side of ∠CBA into line with the other; it is also half the sum of the intercepted arcs. This works whenever B is inside the circle. When B is outside the circle, the rotation associated with the arc closest to B will have the opposite direction, and we get a rotation of magnitude half the measure of the larger arc take away the measure of the smaller arc. For general angles use an algebraic version of the examples done above.
Set I. Note the heavier arcs on the circle in the Danger Angle (17-21). These suggest the right approach. Set II, The Seasons (45-48) are intended to be worked out informally, but they can be dealt with quantitatively as well. See the digression on ancient astronomy below. Angles Problem (56-59). This figure holds the secret to working out the angles in the Tantalus problem, the subject of the Set III exercise on page 97. The person who devised the Tantalus problem removed the clues that link it to this regular 18-sided polygon.
The Set III exercise will become clear if students look at and think about the figure at the bottom right of the page.
▶ A Digression on Ancient Astronomy. In The Seasons exercises students have a chance to think about early efforts to model the movement of the sun in the sky from season to season. This is the “solar theory” of ancient astronomy. Astronomy was a practical matter because the regular annual changes in the sky were a guide to when crops needed to be planted or gathered. Astrology was a strong motivation as well, because the stars and planets were thought to influence human actions.
The Babylonian mathematical astronomy was based on records dating back to the 8th century Bc. Its methods were primarily arithmetic. See, for example, Otto Neugebauer’s The Exact Sciences in Antiquity, available from Dover Publications in 2003, or search for Babylonian astronomy on the Web.
Greek mathematical astronomy used geometric models. ‘These exercises are based on Hipparchus’ model for the movement of the sun along its path (the ecliptic)
through the zodiac. The zodiac is made up of constellations of fixed stars. These are so far away that they represent directions that are unchanged by shifts on the scale of the diagram seen here.
- Given the earth’s position in this figure, point A lies in the direction of the sun relative to earth at the vernal equinox. Viewed from the sun, the equinoxes are the points when the earth’s equatorial plane cuts through the center of the sun. At such a point, the sun is directly overhead at the point on the earth’s equator where it is noon. At this moment, the dividing line between night and day on the earth runs through both poles. Consequently, days and nights are of almost equal duration at this time of year, hence the name equinoz from equal night.
Viewed from the earth, the equinoxes are the times when the sun’s path (the ecliptic) intersects the celestial equator. The celestial equator is the projection of the earth’s equator from the earth’s center onto the celestial sphere (which is simply the set of directions represented by points in the sky, fixed in relationship to the fixed stars). The celestial equator divides the northern and southern hemispheres of the heavens. When the sun stands at the celestial equator as viewed from the earth, it is directly overhead of some point on the earth’s equator (where it is high noon) and the sun illuminates a hemisphere that passes through both poles. This is exactly the configuration described in the previous paragraph, except that we are viewing it from the earth rather than from the sun.
The summer solstice is the instant when the earth’s northern hemisphere is most turned toward the sun. Looking at this from the earth’s point of view, it is when the sun is farthest north on the ecliptic. This time of year gives the longest days in the northern hemisphere. Similarly, the winter solstice is when the sun reaches its furthest south point on the ecliptic and the days are shortest in the northern hemisphere.
In the diagram for Exercises 45-48, the number of degrees from the vernal equinox at A to the summer solstice at B is given as 93.14° & aaa x 360°, where Hipparchus used 94.5 days as the number of days in spring and 365.25 days as the number of days in the year. (Neugebauer in his History of Ancient Mathematical Astronomy, cited earlier, gives Hipparchus’s value of the length of the year more precisely as 3654 — a0 days. His source is Ptolemy in the Almagest.) By a similar calculation, the 92.5 days from the summer solstice at B to the autumnal equinox at C corresponds to the 91.17° of the length of the arc corresponding to summer.
The standard way for the Greeks to deal with circular movements over time was for them to assume that the circular movements were uniform, proceeding at a constant rate. But that does not fit this picture, since ∠AEB = ∠BEC = ∠CED = ∠DEA = 90°. So how
could Hipparchus make a sun moving uniformly around a circle fit into the nonuniform pattern he found for the lengths of the seasons? This is the question students answer here. His solution is simple. He moves the earth off center but keeps the perpendicular directions to the equinozes and solstices the same. This is OK because these directions are tied to points along the zodiac. These points are so far distant that the tiny shift in the earth-sun system leaves them essentially unchanged.
Spring and summer should account for 93.14°+91.17° = 180° + 2 x 2.155°. So we need to move points A and C to new positions A’ and B’, each closer to D by an arc of measure 2.155°. Now if we relocate E to E’ on the segment A’B’, then spring and summer together will have the right duration if the sun moves along the circle at a uniform rate. Next we choose B’ on the major arc from A’ to C’ so that the arc from A’ to A’ has measure 93.14°. This ensures that spring and summer each have the right duration. Finally the foot of the perpendicular from B’ to segment A’B’ gives us the location of M’. This is the construction Neugebauer gives for Hipparchus’ model of the motion of the sun.
If we assume a nominal radius of 60 for the circle here, we find that E’ is displaced 2.2562 units below E and 1.0314 units to the right. The total displacement determined by the Pythagorean Theorem is 2.3891 units, rounded. This displacement makes an angle of −65.4360° with the ray EC in the original picture. If you compute the lengths of the seasons given by this model in degrees, you will find they match those given in the original figure.
These models were simply devices for making predictions that matched observations, particularly the observations made at solstices, equinoxes, conjunctions, eclipses, and other notable events. Just as we may find it easier to think of the sun as going around the earth or of the earth as turning on its axis beneath the sun depending on the problem at hand, the Greeks used first one model, then another. The models are mathematically equivalent, but quite different physically. On page 57 of his History of Ancient Mathematical astronomy, cited earlier, Neugebauer says “The frequent shift from one cinematic model to another is the best indication of the fact that none of these implied that there existed in nature a corresponding mechanical structure.”
Not until Newton do we see a theory of the solar system that attempts to explain the motions of the solar system by a dynamic physical theory. What ancient mathematical astronomy gives is simply a mathematical description of what you will see. It says nothing about the mechanism involved.
Hipparchus knew that to properly track the motions of the sun, you needed to allow for a backward movement of the equinoxes on the ecliptic by 1° per century. This value is close to our modern value for the precession
of the equinoxes. Over the years this adds up. The vernal equinox is known as “the first point of Aries” because it originally took place as the sun entered the constellation of Aries. The vernal equinox now takes place when the sun is in Pisces. In another 600 years it will take place when the sun enters Aquarius (hence, the phrase “the dawning of the Age of Aquarius”). We understand the physical mechanism behind these slow changes. The earth precesses; its axis of rotation turns slowly about an axis perpendicular to the plane of the solar ecliptic like a giant top. Long ago, people knew of these effects, although they had no physical theory of them.
Lesson 6. Tangent Segments and Intersecting Chords. The geometry of eclipses of the sun and of the moon, Exercises 11-15, page 518, is important for the development of geometry and astronomy, and it is discussed at length earlier in these notes, Chapter 11 supplement, “To learn how to calculate better.”
Each of the results in this lesson comes down to noticing that certain products of segments connected with segments related to a given circle must be constant. These products are constant because they are crossed products of terms in a proportion that arises from similar triangles. Given the general idea, you can fiddle around and get whichever result you want.
In the figure for Theorem 67, if you extend AC and DB until they meet at some point Q, then △QBA ≅ △QCD, so QA? QC = QD - QB = QX?’ = 0Q? = r², where QX is tangent to the circle O which runs through A, B, C, and D, and r is the radius of O. The limiting case of the tangent gives us a simple geometric interpretation of the quantity that is constant.
Doing calculations is the best way to drive home these results. Set I has the Lunar Eclipse geometry (11- 15). Note that the distances AS, SE, and EB in the lunar eclipse figure are given in Exercise 53. Set II’s Ridiculous Question is a delight. The answer is obvious, if the question has a definite answer. When the reason this works sinks in, you see why the question is ridiculous. Chord Division (44-52) gives the geometry behind a proposed scheme to measure the curvature of the earth. Points C and D are at the ends of a long canal, whose midpoint is at A. Sighting from C to D the bulge (PA) the water makes in following the earth’s surface at A is measured. We have AP: PB = CP-PD = CP”. The relative error made by assuming the diameter AB is the same length as PB is small, so we may estimate the earth’s radius as Diameter/2 ≅ CP?/2AP. This problem can also be solved using the small angle approximation, treating 0 ≅ sin@ = tan@ as if they are equal for small values of 9. The angle must be expressed in radians to have this work out. This approximation gives the same result as the approximation used above.
Set III will be within the reach of most students.
Summary and Review. To the key points listed here, I would add the symmetries of the circle, reflections in lines through its center or rotations about its center. Often these symmetries give us the keys to answer our questions or solve our problems. Exercises 9-16 are nice because the figure poses the complete exercise. The Golden Ratio (55-62) brings us back to the number ¢ = (1 + V5)/2 again. Algebra shows @ = 1+ 1/¢, from which you can get the continued fraction expansion of ¢. (See, Chapter 14, Lesson 2, Set III in these notes for this fraction.)
Algebra Review. Solving Systems of Equations. This is a nice review of systems of linear equations. The method of using linear combinations of equations generalizes to any number of linear equations in any number of unknowns. If there are more unknowns than there are equations, the set of solutions will form a plane or higher dimensional affine subspace of the space of all possible values for the variables.
We take the equations we are given and combine them in various ways to get conclusions of the sort seen in Example 4, page 528: y = 4.and x = −1. These equations are implied by our original equations and they tell us what x and y must be.