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Chapter 5 : Circles

Chapter 5 : Circles

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Chapter 5 : Circles

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  1. Chapter 5 : Circles • A circle is defined by a center and a radius. Given a point O and a length r, the circle with radius r and center O is defined to be the set of all points A such that OA = r. Much of this chapter will be concentrated with the relationships among circles and various lines. The lines commonly associated with circles are as follows: • A secant is a line that intersects a circle at two points. • A tangent is a line that intersects a circle in one point.

  2. A radius is a line segment with one end point at the center and the other on the circle. Note, as before, that the word “radius” sometimes refers to the length of the segment. • A chord is a line segment with both end points on the circle. • A diameter is a chord that contains the center. • In Fig. 5.1, and are secants, is a tangent, is a radius, and are chords and is a diameter.

  3. We first study intersections of circles and lines. • Lemma. A circle and a line cannot intersect in three or more points. • Proof. The proof will be by contradiction. Assume that we have a circle with center O and radius r, and a line with distinct points A, B, and C on both the line and circle. By the definition of circle, , and all have length r and so .

  4. Hence • and • Comparing the first and third congruences, we see that . . But this contradicts either the exterior angle theorem: Since is an exterior angle of it must be larger than . This contradiction proves the lemma.

  5. Theorem. Let A be a point on a circle with center O, and let be a line through A. Then is tangent to the circle at A if and only if . • Proof. Since this is an “if and only if” theorem, we have two statements to prove: If is perpendicular to at A, then is a tangent, and if is tangent to the circle at A, then .

  6. We first consider the case in which is assumed to be perpendicular to , and we want to show that is a tangent. • So, by way of contradiction we assume that is not tangent to the circle and that there is another point B on the intersection of the circle with . Since A and B are both points on the circle, .

  7. But is a right triangle with hypotenuse . This gives a contradiction because the hypotenuse of a right triangle has to be longer than either of the legs. • Next, to prove the converse, we assume that is not perpendicular to and we will prove that is not tangent to the circle. In order to do this we will find another point in the intersection of the circle with . • Since is not perpendicular to , we construct B on such that is perpendicular to . Then we construct C on (Fig. 5.4) such that . It is not hard to see that . By SAS. Thus, . But since is a radius, this implies that C is also a point on the circle and so completes the proof.

  8. This theorem has a number of consequences. First, we can use it to construct tangents. • Problem. Given a circle C with center O and point A on C, construct a line through A and tangent to C. • Solution. Line will be the line perpendicular to at A. • Corollary. If C is any circle and A is a point on C, then there exists a unique line through A tangent to C. • Both the construction and corollary are easy consequences of the theorem and we omit the proofs.

  9. Theorem. Let C be a circle with center O and let A and B be points on C. Assume P is a point external to C and that . and are tangent to C. Then . • Proof. Consider the triangles and . Each has for one side, by definition of a circle, and and . are each right angles. Hence, by SSA for right triangles. So , as claimed. A P O B

  10. Arcs and Angles • In this section we discuss circular areas and how they are measured. We then prove a theorem relating sizes of arcs to angles inscribed in them. This will prove to be a key result for the rest of our study of circles and also for later chapters. We begin with an easy lemma. • Lemma. Let C and be circles with centers O and and with equal radii. Let A and B be points on C and and be points on . Then if and only if . .

  11. Proof. (Fig. 5.5) Assume , since C and have equal radii, and , then by SSS, and and are corresponding parts of congruent triangles. Conversely, if we assume that . , then by SAS, and here . are corresponding parts of congruent triangles.

  12. Now let C be any circle and A and B be two points on C. A and B divide the circle into two arcs, and, confusingly enough, both arcs are referred to as . • If it is not clear from the context which arc is meant, it is proper to add a point in between, such as or . • Assume, as in Fig. 5.6, that is the smaller of the two. Then we will define the measure of then arc to be that of and the measure to be . In the spirit of the lemma we will say that two arcs and are congruent, written , if they have the same measure and if they come from circles of equal radii (or from the same circles). So if and only if they have the same measure and if the segments and are congruent.

  13. Moreover, in the spirit of Chapter 0, we will often identify an arc with its degree measure. Given three points on a circle, such as A,Q, and B in Fig. 5.6, we will say that is inscribed in the circle and that it subtends arc , to refer to , the arc that does not contain the vertex Q. • Note that in the statement of this theorem we use our notation convention, which identifies angles and arcs with the real numbers that are their degree measures. • Theorem. If we are given a circle with center O and containing points A, B, and C so that subtends the arc , then

  14. Proof. There are three possible cases to consider: Either O lies on a side of , or it lies on the interior of , or it lies outside of . • First, let O be on a side of : say O is on . Consider the triangle . Since it is isosceles, . Now

  15. Hence . Therefore, as claimed. • Next, assume O is inside of as in Fig. 5.8(a). Connect B to O and extend to a point P on the circle so that will be a diameter. We may now apply the previous case to the angles and , since O is on one side of each of them.

  16. Hence and • Adding these two equations yields • The left hand-side of this equation is and the right-hand side is . The last case, in which O is outside [see Fig. 5.8(b)], is similar. The only change is that now . . • Here are some easy consequences of our theorem. • Corollary. If and are each inscribed in a circle and if each subtends the same arc , then • Proof. and .

  17. Corollary. An angle inscribed in a semicircle must be a right angle. • Proof. In this case the arc subtended is . • A third consequence of our theorem is the following slightly more difficult but extremely useful theorem. • Theorem. If we are given a circle with chords and which intersect at a point E inside the circle, then AE . EB = CE . ED.

  18. Proof. The angles and each subtend the arc and therefore are congruent. Likewise, and are congruent, for they each subtend . Hence So • Cross-multiplication now yields the desired result. • In the situation of the theorem and Fig. 5.9. we can also calculate . • Theorem. Given intersecting chords in a circle (as in Fig. 5.9),

  19. Since the sum of the angles of is , we conclude that • Now and . Also, we can express in terms of arcs: • Substituting all of this into the equation for yields as claimed.

  20. Applications to Constructions • We now show how these results can be applied to solve some construction problems. • Problem. Given a circle C with center O and a point P outside of C, find a line that contains P and that is tangent to C. • Solution. Connect and find the midpoint M. Draw a circle with diameter by making M the center and MO = MP the radius. This circle will intersect C at two points, A and B. Both and are solutions in that both are tangent to C.

  21. Proof. How do we prove that and are tangent to C? Recall that, according to a theorem in Section 5.1, we simply need to show that is perpendicular to and is perpendicular to . Now the angle is inscribed in the semicircle with center M, and the angle is inscribed in the semicircle . Hence, each of them is a right angle.

  22. Problem. Given line segments of lengths a and b, construct a line segment of length x so that . In more geometrical terms, construct x such that a square with side x would have area equal to a rectangle with sides of length a and b.

  23. Solution. First, as in Fig. 5.11, construct a line segment of length a + b, with intermediate point E such that AE = a and EB = b. Next construct the midpoint M of . Using M as center, we can now draw a circle with center M and with as diameter. Finally, construct a line perpendicular to through E. This line will intersect the circle at points C and D. Then EC = ED = x. • Proof. Since and are chords intersecting at E, CE . ED = AE . EB = ab. So all we need to show is that CE = ED. Draw and and consider the triangles . and . Since and are radii, , and it is obvious that . Also, and are right angles. So, by SSA for right triangles, and and are corresponding parts of congruent triangles.

  24. Our third construction is related to the problem of solving a quadratic equation. How would you solve an equation such as ? • You could use the quadratic formula and, in principle, now that we know how to take square roots geometrically, we could try to develop a geometric construction based on the quadratic formula. • Another method of solving would be to factor it. We write and try to find two numbers whose sum is 7 and whose product is 12. Of course, 3 and 4 work and they would be the solutions. The geometric problem we will try to solve is to find two segments with sum a and product equal to a given square This corresponds to solving the quadratic equation

  25. Problem. Given line segments of lengths a and b, find two lengths whose sum is a and whose product is . (Or, construct a rectangle with a given area and a given semiperimeter.) • Solution. Let be a line segment of length a, and construct a circle as in Fig. 5.12 with as diameter. Construct a line perpendicular to , at point A. Choose a point C on such that AC = b. Draw a line through C, perpendicular to , and that meets the circle at point D. Finally, drop a perpendicular from D to , meeting at E. Then and are the solutions to the problem.

  26. Proof. We need to calculate the sum and product of AE and EB. First, AE + EB = AB = a. As for the product, let intersect the circle at the second point F. Since and are intersecting chords, we know that AE . EB = DE . EF. To complete the proof we will show that DE = EF = b. is a side of the rectangle DEAC, so DE = AC = b.

  27. To compute EF, let O be the center of the circle and draw the radii and . The triangles and are congruent by SSA for right triangles. Hence FE = DE = b and the proof is complete.

  28. Application to Queen Dido’s Problem • Queen Dido was promised as much land along the coast as she could cover with an ox hide. In order to get as much as possible, she cut and sewed the hide and made it into a long rope. According to the legend, this is how the ancient city of Carthage was found. • From a mathematical point of view, here is the problem Queen Dido faced: She wanted to construct a region such that one side would be a straight line (the seashore), the rest of its boundary would be a fixed length (the length of her rope) and its area would be as great as possible.

  29. Ancient state of North Africa, and at times also the southwestern part of the Mediterranean basin, lasting from about 9th century BCE to 146 BCE. From the 8th century till the 3rd century BCE, Carthage was the dominating power of the western half of the Mediterranean. • The state had its name from the city of Carthage, out on the coast, 10 km from today's Tunis, Tunisia. Carthage had been founded in the 9th century by Phoenician traders of Tyre. Carthage had two first class harbours, and therefore an advantage with the most efficient means of communications of those days, the sea. The Carthaginians soon developed high skills in the building of ships and used this to dominate the seas for centuries. The most important merchandise was silver, lead, ivory and gold, beds and bedding, simple, cheap pottery, jewellery, glassware, wild animals from African, fruit, nuts.

  30. It turns out that the solution to Queen Dido’s problem will be a bit vague on a few technical points. For more details about this as well as other interesting geometric optimization problems, refer to Geometric Inequalities, by Nicholas D. Kazarinoff, in the Mathematical Association of America’s New Mathematical Library, volume 4, 1961. • There are two ingredients to the proof, which we will prove as separate lemmas. • Lemma 1. Of all triangles with BC equal to a given length a and AC equal to a given length b, the triangle of maximum area is the one with .

  31. Lemma 2. Let be fixed segment and let S = the set of all points C such that . Then S is a semicircle with diameter . • We will assume that the two lemmas are true and use them to prove that the semicircle region is the solution to Queen Dido’s problem. Then we will go back and prove the two lemmas. • Proof of Queen Dido’s Theorem. Let us call the region that maximizes the area R and assume that R touches the seashore along the line segment . Let C be any point on the rest of the boundary of R.

  32. We first claim that the triangle is contained in R. The proof of this fact is by contradiction: If the boundary of R sagged in and crossed or , then by pushing it out we could produce a region with an equal perimeter and greater area and this would contradict our definition of R.

  33. Next, we claim that must be a right angle. Again, this proof will be by contradiction. The triangle divides R into three regions (Fig. 5.14) we have labeled 1, 2, and 3. If then we could produce a region with the same perimeter and greater area using lemma 1. If we could push A and B further apart to make . This “pushing” would not affect the lengths of and , so we could still fit regions 1 and 3 together with our new region 2. • In the new figure the area would be greater as guaranteed by lemma 1, and the perimeter would be the same. Since we assumed that R has maximum area, this is impossible and so is not less than a right angle. We can reason to a similar contradiction if we assumed . This forces . , as we claimed.

  34. Now we are done, by use of lemma 2. Our region R has a boundary away from shore that consists of points C on a given side of (the dry side) such that . Hence the boundary of R is a semicircle. • Note that we omitted some technical details. We assumed without proof that the problem has a solution! This means that what we have really proven is that if Queen Dido’s problem has a solution , then the solution is given by a semicircle. • We now backtrack and provide proofs of the lemmas.

  35. Proof of Lemma 1. If then has area . We will show that if then has area less than . If we take as the base, then has area . , where h is the length of the altitude . But is a leg of the right triangle , which has hypotenuse . So or . This implies the area , as claimed. • Proof of Lemma 2. We defined S to be the set Let us now define to be the semicircle with diameter and on the appropriate side.

  36. With this notation we need to show that . To show that two sets are equal we need to show first that if C belongs to , then it belongs to S; then we must show that if C belongs to S, then it belongs to . • If C belongs to then C is a point on the semicircle with diameter . This means that is inscribed in a semicircle, and we know that an angle inscribed in a semicircle must be a right angle. So, by definition, C will be an element of S.

  37. Conversely, now assume that C belongs to S. This means that we are assuming that and we want to show that C is on the semicircle with diameter . Our proof will be by contradiction. If C is not on this semicircle, then there is another point where meets the semicircle. As before, is a right angle because it is inscribed in a semicircle. Now we can get a contradiction if we consider . This triangle has two right angles, at C and at . This is impossible, and this completes the proof.

  38. More on Arcs ands Angles • We proved various theorems concerning chords before. In Now, we will prove analogous theorems for secants and tangents. • Theorem. If B, C, D, and E are points on a circle such that . and intersect at a point A outside of the circle (as in Fig. 5.17), then • (a) • (b)

  39. Proof. (a): Consider . But and Also, • Now, by substitution,

  40. (b) For this half of the theorem we consider the triangles and (See Fig. 5.18). Each has for one angle. Also, since each subtends the arc . Thus, by AA, . Hence . If we cross multiply we get (b). • We now turn to the case of tangents. In order to repeat the proof from the secant case, we need this preliminary result.

  41. Theorem. Assume at a point A on the circle that there is a chord and a tangent . Then • We remark that the line segment and the line make two different angles with each other (they are supplementary) and that there are two arcs that could be labeled . Each angle is half of the arc that it cuts off-the larger angle corresponds to the larger arc and the smaller angle corresponds to the smaller arc. • Proof. Assume that C is such that is less than or equal to , as in the diagram. Let O be the center of the circle and extend to a diameter .

  42. Then is perpendicular to and • But and . The result now follows by subtraction. The proof in the case of obtuse is the same, except that will be rather than

  43. With this tool, the following theorems can be proved easily. • Theorem. If B, C, and D are points on a circle such that the tangent at B and the secant intersect at a point A, then (a) (b)

  44. Theorem. Suppose B and C are points on a circle such that the tangent at B and the tangent at C meet at a point A. Let P and Q be points on the circle such that . Then B A Q P C