GEOMETRY OF THE CIRCLE

Transcription

1 HTR GMTRY F TH IRL arly geometers in many parts of the world knew that, for all circles, the ratio of the circumference of a circle to its diameter was a constant. Today, we write d 5p, but early geometers did not use the symbol p to represent this constant. uclid established that the ratio of the area of a circle to the square of its diameter was also a constant, that is, 5 k. How do these d 2 constants, p and k, relate to one another? rchimedes ( ) proposed that the area of a circle was equal to the area of a right triangle whose legs have lengths equal to the radius, r, and the circumference,, of a circle. Thus 2r. He used indirect proof and the areas of inscribed and circumscribed polygons to prove his conjecture and to prove that 3 0 7,p,3 7. Since this inequality can be written as p ,rchimedes approximation was correct to two decimal places. Use rchimedes formula for the area of a circle and the facts that d 5pand d 2r to show that pr 2 and that 4k = p. 3 HTR TL F NTNTS 3- rcs and ngles 3-2 rcs and hords 3-3 Inscribed ngles and Their Measures 3-4 Tangents and Secants 3-5 ngles Formed by Tangents, hords, and Secants 3-6 Measures of Tangent Segments, hords, and Secant Segments 3-7 ircles in the oordinate lane 3-8 Tangents and Secants in the oordinate lane hapter Summary Vocabulary Review xercises umulative Review 535

2 536 Geometry of the ircle 3- RS N NGLS In hapter, we defined a sphere and found that the intersection of a plane and a sphere was a circle. In this chapter, we will prove some important relationships involving the measures of angles and line segments associated with circles. Recall the definition of a circle. FINITIN circle is the set of all points in a plane that are equidistant from a fixed point of the plane called the center of the circle. If the center of a circle is point, the circle is called circle, written in symbols as (. radius of a circle (plural, radii) is a line segment from the center of the circle to any point of the circle. The term radius is used to mean both the line segment and the length of the line segment. If,, and are points of circle, then,, and are radii of the circle. Since the definition of a circle states that all points of the circle are equidistant from its center,.thus, > > because equal line segments are congruent. We can state what we have just proved as a theorem. Theorem 3. ll radii of the same circle are congruent. circle separates a plane into three sets of points. If we let the length of the radius of circle be r, then: r oint is on the circle if r. oint is outside the circle if r. oint is inside the circle if r. The interior of a circle is the set of all points whose distance from the center of the circle is less than the length of the radius of the circle. The exterior of a circle is the set of all points whose distance from the center of the circle is greater than the length of the radius of the circle. entral ngles Recall that an angle is the union of two rays having a common endpoint and that the common endpoint is called the vertex of the angle. FINITIN central angle of a circle is an angle whose vertex is the center of the circle.

3 rcs and ngles 537 In the diagram, LM and MR are central angles because the vertex of each angle is point, the center of the circle. L R M Types of rcs n arc of a circle is the part of the circle between two points on the circle. In the diagram,,,, and are points on circle and intersects the circle at two distinct points, and, separating the circle into two arcs. X Minor arc ( X ) Major arc ( ). If m 80, points and and the points of the circle in the interior of make up minor arc, written as X. 2. oints and and the points of the circle not in the interior of make up major arc. major arc is usually named by three points: the two endpoints and any other point on the major arc. Thus, the major arc with endpoints and is written as X or X. 3. If m 80, points and separate circle into two equal parts, each of which is called a semicircle. In the diagram above, X and X name two different semicircles. egree Measure of an rc n arc of a circle is called an intercepted arc, or an arc intercepted by an angle, if each endpoint of the arc is on a different ray of the angle and the other points of the arc are in the interior of the angle. FINITIN The degree measure of an arc is equal to the measure of the central angle that intercepts the arc.

4 538 Geometry of the ircle F G In circle, G is a straight angle, m G 80, and m FG 80. Therefore, the degree measure of FG X is 80, written as mfgx lso, since mgf X 5 80, mf X mfg X and Therefore, mgfx mfg X The degree measure of a semicircle is 80. Thus: 2. The degree measure of a major arc is equal to 360 minus the degree measure of the minor arc having the same endpoints. o not confuse the degree measure of an arc with the length of an arc. The degree measure of every circle is 360 but the circumference of a circle is 2p times the radius of the circle. xample: in circle, cm, and in circle,.5 cm. In both circles, the degree measure of the circle is 360 but the circumference of circle is 2p centimeters, and the circumference of circle is 3p centimeters. ongruent ircles, ongruent rcs, and rc ddition FINITIN ongruent circles are circles with congruent radii. If > rr, then circles and are congruent. FINITIN ongruent arcs are arcs of the same circle or of congruent circles that are equal in measure.

5 If ( > (r and mx 5 mrr 60, then X > rr. However, if circle is not congruent to circle, then X is not congruent to ss X even if X X X and ss have the same degree measure. rcs and ngles 539 X ostulate 3. rc ddition ostulate If X and X are two arcs of the same circle having a common endpoint and no other points in common, then X X X and mx mx m X. The arc that is the sum of two arcs may be a minor arc, a major arc, or semicircle. For example,,,, and are points of circle, m X 5 90, mx 5 40, and h and h are opposite rays.. Minor arc: X X X 40 lso, 90 mx 5 m X m X 2. Major arc: X = lso, m X 5 m X m X X X 3. Semicircle: Since h and are opposite rays, is a straight angle. Thus, X h X X, a semicircle. lso, mx m X 5 m X 5 80.

6 540 Geometry of the ircle Theorem 3.2a In a circle or in congruent circles, if central angles are congruent, then their intercepted arcs are congruent Given ircle circle,, and. rove X > X and X > rr X. roof It is given that and. Therefore, m m m because congruent angles have equal measures. Then since the degree measure of an arc is equal to the degree measure of the central angle that intercepts that arc, mx 5 m X 5 mrr. It is also given that circle and circle are congruent circles. ongruent arcs are defined as arcs of the same circle or of congruent circles that are equal in measure. Therefore, since their measures are equal, X > X and X > rr. X X The converse of this theorem can be proved by using the same definitions and postulates. Theorem 3.2b In a circle or in congruent circles, central angles are congruent if their intercepted arcs are congruent. Theorems 3.2a and 3.2b can be written as a biconditional. Theorem 3.2 In a circle or in congruent circles, central angles are congruent if and only if their intercepted arcs are congruent. XML Let h and h be opposite rays and m 75. Find: a. m b. mx c. mx d. mx e. m X Solution a. m/ 5 m/ 2 m/

7 rcs and ngles 54 b. mx m 75 c. mx m 05 d. mx m 80 e. m X 5 m X m X or m X m X nswers a. 05 b. 75 c. 05 d. 80 e. 255 xercises Writing bout Mathematics. Kay said that if two lines intersect at the center of a circle, then they intercept two pairs of congruent arcs. o you agree with Kay? Justify your answer. 2. Four points on a circle separate the circle into four congruent arcs: X, X, X, and X. Is it true that g ' g? Justify your answer. eveloping Skills In 3 7, find the measure of the central angle that intercepts an arc with the given degree measure In 8 2, find the measure of the arc intercepted by a central angle with the given measure In 3 22, the endpoints of are on circle, m 89, and m 42. Find each measure. 3. m 4. mx 5. mx 6. m 7. mx 8. m m X 20. m X 2. m 22. m X

8 542 Geometry of the ircle In 23 32,, Q, S, and R are points on circle,m Q 00, m QS 0, and m SR 35. Find each measure. 23. mqx 24. mqs X 25. msrx 26. m R mqs X mrx R m QR 30. mqsr X S 0 Q 3. msr X 32. mrq X pplying Skills 33. Given: ircle with X > X. rove: 34. Given: and intersect at, and the endpoints of and are on circle. rove: > 35. In circle, '. Find mx and m. 36. oints,,, and lie on circle, and ' at. rove that quadrilateral is a square. X Hands-n ctivity For this activity, you may use compass, protractor, and straightedge, or geometry software.. raw circle with a radius of 2 inches and circle with a radius of 3 inches. 2. raw points and on the circle so that mx 60 and points and on circle so that mrr X Show that.

9 rcs and hords RS N HRS FINITIN chord of a circle is a line segment whose endpoints are points of the circle. diameter of a circle is a chord that has the center of the circle as one of its points. In the diagram, and are chords of circle. Since is a point of, is a diameter. Since and are the lengths of the radius of circle, and is the midpoint of. If the length of the radius of circle is r, and the length of the diameter is d, then d r r 5 2r That is: d 2r The endpoints of a chord are points on a circle and, therefore, determine two arcs of a circle, a minor arc and a major arc. In the diagram, chord, central, minor X, and major X are all determined by points and. We proved in the previous section that in a circle, congruent central angles intercept congruent arcs. Now we can prove that in a circle, congruent central angles have congruent chords and that congruent arcs have congruent chords. Theorem 3.3a In a circle or in congruent circles, congruent central angles have congruent chords. Given ( > (r and rove > > rr

10 544 Geometry of the ircle roof We will show that by SS. It is given that and. Therefore, by the transitive property of congruence. Since,,,, rr, and rr are the radii of congruent circles, these segments are all congruent: > > > > rr > rr Therefore, by SS,. Since corresponding parts of congruent triangles are congruent, > > rr. The converse of this theorem is also true. Theorem 3.3b In a circle or in congruent circles, congruent chords have congruent central angles. Given ( > (r and > > rr rove Strategy This theorem can be proved in a manner similar to Theorem 3.3a: prove that by SSS. The proof of Theorem 3.3b is left to the student. (See exercise 23.) Theorems 3.3a and 3.3b can be stated as a biconditional. Theorem 3.3 In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. Since central angles and their intercepted arcs have equal degree measures, we can also prove the following theorems. Theorem 3.4a In a circle or in congruent circles, congruent arcs have congruent chords. Given rove ( > (r and > > rr X > X > rr X

11 rcs and hords 545 roof First draw line segments from to,,,,, and. ongruent arcs have congruent central angles. Therefore,. In a circle or in congruent circles, two chords are congruent if and only if their central angles are congruent. Therefore, > > rr. The converse of this theorem is also true. Theorem 3.4b In a circle or in congruent circles, congruent chords have congruent arcs. Given rove Strategy ( > (r and > > rr X X > X > rr First draw line segments from to,,,, and. rove by SSS. Then use congruent central angles to prove congruent arcs. The proof of Theorem 3.4b is left to the student. (See exercise 24.) Theorems 3.4a and 3.4b can be stated as a biconditional. Theorem 3.4 In a circle or in congruent circles, two chords are congruent if and only if their arcs are congruent. XML In circle, mx 35, m 0, and is a diameter. a. Find mx and mx. b. xplain why =. Solution a. mx 5 m/ 5 0 mx m X 2 m X b. In a circle, arcs with equal measure are congruent. Therefore, since mx 35 and mx 35, X X. In a circle, chords are congruent if their arcs are congruent. Therefore, > and =.

12 546 Geometry of the ircle hords quidistant from the enter of a ircle We defined the distance from a point to a line as the length of the perpendicular from the point to the line.the perpendicular is the shortest line segment that can be drawn from a point to a line. These facts can be used to prove the following theorem. Theorem 3.5 diameter perpendicular to a chord bisects the chord and its arcs. Given iameter of circle, chord, and ' at. rove >, X > X, and X > X. roof Statements Reasons. raw and.. Two points determine a line. 2. ' 2. Given. 3. and are right 3. erpendicular lines intersect to angles. form right angles. 4. > 4. Radii of a circle are congruent. 5. > 5. Reflexive property of congruence HL. 7. > 7. orresponding parts of congruent triangles are congruent orresponding parts of congruent triangles are congruent. 9. X > X 9. In a circle, congruent central angles have congruent arcs. 0. is the supplement of 0. If two angles form a linear pair,. then they are supplementary. is the supplement of... Supplements of congruent angles are congruent. 2. X > X 2. In a circle, congruent central angles have congruent arcs.

13 rcs and hords 547 Since a diameter is a segment of a line, the following corollary is also true: orollary 3.5a Theorem 3.6 line through the center of a circle that is perpendicular to a chord bisects the chord and its arcs. n apothem of a circle is a perpendicular line segment from the center of a circle to the midpoint of a chord. The term apothem also refers to the length of the segment. In the diagram, is the midpoint of chord in circle, ', and, or, is the apothem. The perpendicular bisector of the chord of a circle contains the center of the circle. Given ircle, and chord with midpoint M and perpendicular bisector k. M rove oint is a point on k. roof In the diagram, M is the midpoint of chord in circle. k Then, M = M and = (since these are radii). oints and M are each equidistant from the endpoints of. Two points that are each equidistant from the endpoints of a line segment determine the perpendicular bisector of the line segment. Therefore, M g is the perpendicular bisector of. Through a point on a line there is only one perpendicular line. Thus, M g and k are the same line, and is on k, the perpendicular bisector of. Theorem 3.7a If two chords of a circle are congruent, then they are equidistant from the center of the circle. Given ircle with >, ', and F '. rove roof > F F line through the center of a circle that is perpendicular to a chord bisects the chord and its arcs. Therefore, g and F g bisect the congruent chords and. Since halves of congruent segments are congruent, > F. raw and. Since and are radii of the same circle, >. Therefore, F by HL, and > F.

14 548 Geometry of the ircle The converse of this theorem is also true. Theorem 3.7b If two chords of a circle are equidistant from the center of the circle, then the chords are congruent. Given ircle with ', F ', and > F. rove > F roof raw and. Since and are radii of the same circle, >. Therefore, F by HL, and > F. line through the center of a circle that is perpendicular to a chord bisects the chord. Thus, > and F > F. Since doubles of congruent segments are congruent, >. We can state theorems 3.7a and 3.7b as a biconditional. Theorem 3.7 Two chords are equidistant from the center of a circle if and only if the chords are congruent. What if two chords are not equidistant from the center of a circle? Which is the longer chord? We know that a diameter contains the center of the circle and is the longest chord of the circle. This suggests that the shorter chord is farther from the center of the circle. () Let and be two chords of circle and. (2) raw ' and F '. (3) line through the center of a circle that is perpendicular to the chord bisects the chord. F Therefore, 2 5 and 2 5 F. (4) The distance from a point to a line is the length of the perpendicular from the point to the line. Therefore, is the distance from the center of the circle to, and F is the distance from the center of the circle to. Recall that the squares of equal quantities are equal, that the positive square roots of equal quantities are equal, and that when an inequality is multiplied by a negative number, the inequality is reversed.

15 rcs and hords 549 (5) Since :, 2, 2, F 2, F F 2 (6) Since and F are right triangles and = : F 2 F 2 (7) When equal quantities are added to both sides of an inequality, the order of the inequality remains the same. Therefore, adding the equal quantities from step 6 to the inequality in step 5 gives: F 2 F 2 2 F 2 2. F 2. F Therefore, the shorter chord is farther from the center of the circle.we have just proved the following theorem: Theorem 3.8 In a circle, if the lengths of two chords are unequal, then the shorter chord is farther from the center. XML 2 In circle, mx 90 and 6. a. rove that is a right triangle. b. Find. c. Find, the apothem to. Solution a. If mx 90, then m 90 because the measure of an arc is equal to the measure of the central angle that intercepts the arc. Since is a right angle, is a right triangle. 6 90

16 550 Geometry of the ircle 90 6 b. Use the ythagorean Theorem for right. Since and are radii, "72 5 "36"2 5 6"2 c. Since is the apothem to, ' and bisects. Therefore, 5 3"2. In right, " "8 5 "9"2 5 3"2 Note: In the example, since is an isosceles right triangle, m is 45. Therefore is also an isosceles right triangle and. olygons Inscribed in a ircle If all of the vertices of a polygon are points of a circle, then the polygon is said to be inscribed in the circle. We can also say that the circle is circumscribed about the polygon. In the diagram:. olygon is inscribed in circle. 2. ircle is circumscribed about polygon. In an earlier chapter we proved that the perpendicular bisectors of the sides of a triangle meet at a point and that that point is equidistant from the vertices of the triangle. In the diagram, L g, M g, and N g are the perpendicular bisectors of the sides of. very point on the perpendicular bisectors of a line segment is equidistant from the endpoints of the line segment. Therefore, and,, and are points on a circle with center at, that is, any triangle can be inscribed in a circle. N L M

17 rcs and hords 55 XML 3 roof rove that any rectangle can be inscribed in a circle. Let be any rectangle. The diagonals of a rectangle are congruent, so > and. Since a rectangle is a parallelogram, the diagonals of a rectangle bisect each other. If and intersect at, then 2 and 2. Halves of equal quantities are equal. Therefore, and the vertices of the rectangle are equidistant from. Let be the center of a circle with radius. The vertices of are on the circle and is inscribed in the circle. xercises Writing bout Mathematics. aniela said that if a chord is 3 inches from the center of a circle that has a radius of 5 inches, then a right triangle is formed by the chord, its apothem, and a radius. dditionally, the length of the chord is 4 inches. o you agree with aniela? xplain why or why not. 2. Two angles that have the same measure are always congruent. re two arcs that have the same measure always congruent? xplain why or why not. eveloping Skills In 3 7, find the length of the radius of a circle whose diameter has the given measure in cm 5. 3 ft mm 7. "24 cm In 8 2, find the length of the diameter of a circle whose radius has the given measure in ft 0. 7 cm. 6.2 mm 2. "5 yd 3. In circle, is a diameter, 3x 3, and 2x 5. Find the length of the radius and of the diameter of the circle. In 4 2, is a diameter of circle, is a chord of the circle, and ' at. 4. If 8 and 3, find. 5. If 48 and 7, find. 6. If 20 and 25, find. 7. If 2 and 8, find.

18 552 Geometry of the ircle 8. If 8 and 5, find. 9. If 20 and 5, find. 20. If m 90, and 30"2, find 2. If m 60, and 30, find and. and. 22. In circle, chord LM is 3 centimeters from the center and chord RS is 5 centimeters from the center. Which is the longer chord? pplying Skills 23. rove Theorem 3.3b, In a circle or in congruent circles, congruent chords have congruent central angles. 24. rove Theorem 3.4b, In a circle or in congruent circles, congruent chords have congruent arcs. 25. iameter of circle intersects chord at and bisects X at. rove that bisects chord and is perpendicular to chord. 26. The radius of a spherical ball is 3 centimeters. piece that has a plane surface is cut off of the ball at a distance of 2 centimeters from the center of the ball. What is the radius of the circular faces of the cut pieces? 27. Triangle is inscribed in circle. The distance from the center of the circle to is greater than the distance from the center of the circle to, and the distance from the center of the circle to is greater than the distance from the center of the circle to. Which is the largest angle of? Justify your answer. 3-3 INSRI NGLS N THIR MSURS In the diagram, is an angle formed by two chords that have a common endpoint on the circle. FINITIN n inscribed angle of a circle is an angle whose vertex is on the circle and whose sides contain chords of the circle. We can use the fact that the measure of a central angle is equal to the measure of its arc to find the relationship between and the measure of its arc, X.

19 Inscribed ngles and Their Measures 553 x x 2x 2x S ne of the sides of the inscribed angle contains a diameter of the circle. onsider first an inscribed angle,, with a diameter of circle. raw. Then is an isosceles triangle and m m x. ngle is an exterior angle and m x x 2x. Since is a central angle, m mx 5 2x. Therefore, m x X 2 m. We have shown that when one of the sides of an inscribed angle contains a diameter of the circle, the measure of the inscribed angle is equal to one-half the measure of its intercepted arc. Is this true for angles whose sides do not contain the center of the circle? S 2 The center of the circle is in the interior of the angle. Let be an inscribed angle in which the center of the circle is in the interior of the angle. raw, a diameter of the circle. Then: m X and m X 2 m 2 m Therefore: m/ 5 m/ m/ 5 X 2 m 2 mx 5 X 2 (m m X ) 5 X 2 m S 3 The center of the circle is not in the interior of the angle. Let be an inscribed angle in which the center of the circle is not in the interior of the angle. raw, a diameter of the circle. Then: m X and m X 2 m 2 m Therefore: m/ 5 m/ 2 m/ 5 X 2 m 2 2 mx 5 X 2 (m 2 m X ) 5 X 2 m These three possible positions of the sides of the circle with respect to the center the circle prove the following theorem: Theorem 3.9 The measure of an inscribed angle of a circle is equal to one-half the measure of its intercepted arc. There are two statements that can be derived from this theorem.

20 554 Geometry of the ircle orollary 3.9a n angle inscribed in a semicircle is a right angle. roof: In the diagram, is a diameter of circle, and is inscribed in semicircle. lso is a semicircle whose degree measure is 80. Therefore, m 5 2 m 5 2 (80) 5 90 X X X Since any triangle can be inscribed in a circle, the hypotenuse of a triangle can be the diameter of a circle with the midpoint of the hypotenuse the center of the circle. orollary 3.9b If two inscribed angles of a circle intercept the same arc, then they are congruent. roof: In the diagram, and are inscribed angles and each angle intercepts X. Therefore, m X 2 and m X m 2 m. Since and have equal measures, they are congruent. XML Solution Triangle is inscribed in circle,m 70, and mx a. mx b. m c. m d. mx a. If the measure of an inscribed angle is one-half the measure of its intercepted arc, then the measure of the intercepted arc is twice the measure of the inscribed angle. m/ 5 X 2 2m/ 5 mx m 2(70) 5 mx 40 5 mx 00. Find: 70 00

21 b. m/ 5 X 2 m c. m/ (m/ m/) 5 2 (00) (50 70) d mx 5 2m/ (60) 5 20 nswers a. 40 b. 50 c. 60 d. 20 Inscribed ngles and Their Measures SUMMRY Type of ngle egree Measure xample entral ngle The measure of a central angle is equal to the measure of its intercepted arc. Inscribed ngle The measure of an inscribed angle is equal to one-half the measure of its intercepted arc. m mx m/ 5 2 m X xercises Writing bout Mathematics. xplain how you could use orollary 3.9a to construct a right triangle with two given line segments as the hypotenuse and one leg.

22 556 Geometry of the ircle 2. In circle, is an inscribed angle and mx 50. In circle, QR is an inscribed angle and mrx 50. Is QR if the circles are not congruent circles? Justify your answer. eveloping Skills In 3 7, is a point on circle not on X, an arc of circle. Find m for each given m X In 8 2, is a point on circle not on X, an arc of circle. Find mx for each given m Triangle is inscribed in a circle, m 80 and mx 88. Find: a. mx b. m c. m d. mx e. m X Triangle F is inscribed in a circle, > F, and mfx 00. Find: a. m b. mx c. m F d. m e. mfx F 00 In 5 7, chords and intersect at in circle. 5. If m 42 and m 04, find: a. m b. mx c. mx d. m e. m 6. If and m 40, find: a. m b. mx c. mx d. m e. m 7. If mx 00, mx 0, and mx 96, find: a. mx b. m c. m d. m e. m 8. Triangle is inscribed in a circle and mx : mx : mx 2 : 3 : 7. Find: a. mx b. mx c. mx d. m e. m f. m

23 Inscribed ngles and Their Measures 557 X 5 mst X 5 mtr X 9. Triangle RST is inscribed in a circle and mrs. Find: a. mrs X b. mstx c. mtrx d. m R e. m S f. m T pplying Skills 20. In circle, LM and RS intersect at. a. rove that LR SM. b. If L 5 cm, R 2 cm, and S 0 cm, find M. L R M S 2. Triangle is inscribed in a circle. If mx 00 and mx 30, prove that is isosceles. 22. arallelogram is inscribed in a circle. X X a. xplain why m = m. X b. Find m and m. X c. xplain why parallelogram must be a rectangle. G F x. 23 x. 24 x Triangle F is inscribed in a circle and G is any point not on F. If mx mf X 5 mfg, show that F is a right triangle. 24. In circle, and are diameters. If >, prove that by S. 25. hords and of circle intersect at. If X X, prove that. 26. rove that a trapezoid inscribed in a circle is isosceles. X X X 27. Minor and major are arcs of circle and. rove that X > X. 28. Minor X and major X are arcs of circle and X > X. rove that. 29. In circle, and are diameters. rove that. X

24 558 Geometry of the ircle 30. oints,,,,, and F are on circle, F, and. and F are trapezoids. rove that X > X > F X > X. 3. Quadrilateral is inscribed in circle, and X is not congruent to X. rove that is not a parallelogram. F 3-4 TNGNTS N SNTS In the diagram, line p has no points in common with the circle. Line m has one point in common with the circle. Line m is said to be tangent to the circle. Line k has two points in common with the circle. Line k is said to be a secant of the circle. k p m FINITIN tangent to a circle is a line in the plane of the circle that intersects the circle in one and only one point. FINITIN secant of a circle is a line that intersects the circle in two points. Let us begin by assuming that at every point on a circle, there exists exactly one tangent line. We can state this as a postulate. ostulate 3.2 t a given point on a given circle, one and only one line can be drawn that is tangent to the circle. Let be any point on circle and be a radius to that point. If line m containing points and Q is perpendicular to, then Q because the perpendicular is the shortest distance from a point to a line. Therefore, every point on the line except is outside of circle and line m must be tangent to the circle. This establishes the truth of the following theorem. Q m

25 Tangents and Secants 559 Theorem 3.0a If a line is perpendicular to a radius at a point on the circle, then the line is tangent to the circle. The converse of this theorem is also true. Theorem 3.0b If a line is tangent to a circle, then it is perpendicular to a radius at a point on the circle. Given Line m is tangent to circle at. rove Line m is perpendicular to. m b roof We can use an indirect proof. ssume that m is not perpendicular to. Then there is some line b that is perpendicular to at since, at a given point on a given line, one and only one line can be drawn perpendicular to the given line. Then by Theorem 3.0a, b is a tangent to circle at. ut this contradicts the postulate that states that at a given point on a circle, one and only one tangent can be drawn. Therefore, our assumption is false and its negation must be true. Line m is perpendicular to. We can state Theorems 3.0a and 3.0b as a biconditional. Theorem 3.0 line is tangent to a circle if and only if it is perpendicular to a radius at its point of intersection with the circle. ommon Tangents FINITIN common tangent is a line that is tangent to each of two circles. In the diagram, g is tangent to circle at and to circle at.tangent g is said to be a common internal tangent because the tangent intersects the line segment joining the centers of the circles.

26 560 Geometry of the ircle In the diagram, g is tangent to circle at and to circle at. Tangent g is said to be a common external tangent because the tangent does not intersect the line segment joining the centers of the circles. The diagrams below show that two circles can have four, three, two, one, or no common tangents. 4 common tangents 3 common tangents 2 common tangents common tangent No common tangents T S Two circles are said to be tangent to each other if they are tangent to the same line at the same point. In the diagram, ST g is tangent to circle and to circle at T. ircles and are tangent externally because every point of one of the circles, except the point of tangency, is an external point of the other circle. In the diagram, MN g is tangent to circle and to circle at M. ircles and are tangent internally because every point of one of the circles, except the point of tangency, is an internal M point of the other circle. N XML Given: ircles and with a common internal tangent, g, tangent to circle at and circle at, and the intersection of r and g. rove: 5 r

27 Tangents and Secants 56 roof We will use similar triangles to prove the segments proportional. Line g is tangent to circle at and circle at. line tangent to a circle is perpendicular to a radius drawn to the point of tangency. Since perpendicular lines intersect to form right angles and all right angle are congruent,. lso, because vertical angles are congruent. Therefore, by. The lengths of corresponding sides of similar triangles are proportional. Therefore, 5 r. XML 2 ircle is tangent to g at, is tangent to g at, and r intersects g at. a. rove that 5 r. b. If 8, 2, and 9, find. Solution a. We know that because they are right angles and that because they are vertical angles. Therefore, by and 5 r. b r r 9 8r 5 36 r nswer Tangent Segments FINITIN tangent segment is a segment of a tangent line, one of whose endpoints is the point of tangency. In the diagram, Q and R are tangent segments of the tangents Q g and R g to circle from. Q R Theorem 3. Tangent segments drawn to a circle from an external point are congruent.

28 562 Geometry of the ircle Given Q g tangent to circle at Q and R g tangent to circle at R. Q rove Q > R R roof raw Q, R, and. Since Q and R are both radii of the same circle, Q > R. Since Q and R are tangent to the circle at Q and R, Q and R are both right angles, and Q and R are right triangles. Then is the hypotenuse of both Q and R. Therefore, Q R by HL. orresponding parts of congruent triangles are congruent, so Q > R. The following corollaries are also true. orollary 3.a If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. Given Q g tangent to circle at Q and R g tangent to circle at R. Q rove Strategy h bisects RQ. Use the proof of Theorem 3. to show that angles Q and R are congruent. R orollary 3.b If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency. Given Q g tangent to circle at Q and R g tangent to circle at R. Q rove Strategy h bisects QR. Use the proof of Theorem 3. to show that angles Q and R are congruent. R

29 Tangents and Secants 563 The proofs of orollaries 3.a and 3.b are left to the student. (See exercises 5 and 6.) olygon ircumscribed bout a ircle polygon is circumscribed about a circle if each side of the polygon is tangent to the circle.when a polygon is circumscribed about a circle, we also say that the circle is inscribed in the polygon. For example, in the diagram, is tangent to circle at, is tangent to circle at F, is tangent to circle at G, is tangent to circle at H. Therefore, is circumscribed about circle and circle is inscribed in quadrilateral. If is circumscribed about circle, then we know that,, and are the bisectors of the angles of, and is the point at which the angle bisectors of the angles of a triangle intersect. F F H G XML 3 Solution,, and are tangent to circle at,, and F, respectively. If F 6, 7, and 5, find the perimeter of. Tangent segments drawn to a circle from an external point are congruent. F 6 7 F 5 Therefore, F F F erimeter nswer

30 564 Geometry of the ircle XML 4 oint is a point on a line that is tangent to circle at R, is 2.0 centimeters from the center of the circle, and the length of the tangent segment from is 8.0 centimeters. a. Find the exact length of the radius of the circle. b. Find the length of the radius to the nearest tenth. Solution a. is 2 centimeters from the center of circle ; 2. The length of the tangent segment is 8 centimeters; R 8. line tangent to a circle is perpendicular to the radius drawn to the point of tangency; R is a right triangle. R 8 cm 2 cm R 2 R "80 5 "6"5 5 4"5 b. Use a calculator to evaluate 4"5. NTR: 4 2nd 5 NTR ISLY: To the nearest tenth, 8.9. nswers a. 4 "5 cm b. 8.9 cm xercises Writing bout Mathematics. Line l is tangent to circle at and line m is tangent to circle at. If is a diameter, does l intersect m? Justify your answer. 2. xplain the difference between a polygon inscribed in a circle and a circle inscribed in a polygon.

31 eveloping Skills In 3 and 4, is circumscribed about circle and,, and F are points of tangency. 3. If 5, 5, and F 0, find the lengths of the sides of the triangle and show that the triangle is isosceles. 4. If F 0, 20, and 30, find the lengths of the sides of the triangle and show that the triangle is a right triangle. In 5, Q is tangent to circle at, SQ is tangent to circle at S, and Q intersects circle at T and R. 5. If 5 and Q 20, find: a. Q b. SQ c. TQ 6. If Q 25 and Q 24, find: a. b. RT c. RQ 7. If 0 and Q 26, find: a. Q b. RQ c. TQ 8. If 6 and TQ 3, find: a. Q b. Q c. SQ 9. If S 9 and RQ 32, find: a. Q b. SQ c. Q 0. If Q 3x, SQ 5x 8, and S x, find: a. Q b. SQ c. S d. Q. If SQ 2x, S 2x 2, and Q 3x, find: a. x b. SQ c. S d. Q 2. The sides of are tangent to a circle at,, and F. If 4, 7, and the perimeter of the triangle is 30, find: a. b. c. F d. F e. f. Tangents and Secants 565 T F S R Q 3. Line R g is tangent to circle at and R intersects the circle at M, the midpoint of R. If R 3.00 cm, find the length of the radius of the circle: a. in radical form b. to the nearest hundredth M R F 4. oints, F, G, and H are the points of tangency to circle of,,, and, respectively. The measure of FX is 80, of FGX is 70, and of GHX is 50. Find: a. m F b. m FG c. m GH d. m H e. m f. m g. m H h. m F i. m GF j. m HG k. the sum of the measures of the angles of quadrilateral F G H

32 566 Geometry of the ircle pplying Skills 5. rove orollary 3.a, If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle formed by the tangents. 6. rove orollary 3.b, If two tangents are drawn to a circle from an external point, then the line segment from the center of the circle to the external point bisects the angle whose vertex is the center of the circle and whose rays are the two radii drawn to the points of tangency. 7. Lines Q g and R g are tangent to circle at Q and R. a. rove that QR RQ. b. raw intersecting RQ at S and prove that QS RS Q and ' QR. R c. If 0, SQ 4 and S S, find S and S. 8. Tangents and to circle are perpendicular to each other at. rove: a. > b. "2 c. is a square. 9. Isosceles is circumscribed about circle. The points of tangency of the legs, and, are and F, and the point of tangency of the base,, is. rove that is the midpoint of. 20. Line g is a common external tangent to circle and circle. g is tangent to circle at and to circle at, and. g a. rove that r is not parallel to g. g b. Let be the intersection of r and g. rove that. c. If r 5 2 3, and 2, find,,,, and. g 2. Line is a common internal tangent to circles and. g is tangent to circle at and to circle at, and. The intersection of r and g is. a. rove that. b. rove that.

33 Hands-n ctivity onsider any regular polygon. onstruct the angle bisectors of each interior angle. Since the interior angles are all congruent, the angles formed are all congruent. Since the sides of the regular polygon are all congruent, congruent isosceles triangles are formed by S. ny two adjacent triangles share a common leg. Therefore, they all share the same vertex. Since the legs of the triangles formed are all congruent, the vertex is equidistant from the vertices of the regular polygon. This common vertex is the center of the regular polygon. ngles Formed by Tangents, hords, and Secants 567 In this Hands-n ctivity, we will use the center of a regular polygon to inscribe a circle in the polygon. a. Using geometry software or compass, protractor, and straightedge, construct a square, a regular pentagon, and a regular hexagon. For each figure: () onstruct the center of the regular polygon. (The center is the intersection of the angle bisectors of a regular polygon.) (2) onstruct an apothem or perpendicular from to one of the sides of regular polygon. (3) onstruct a circle with center and radius equal to the length of the apothem. b. rove that the circles constructed in part a are inscribed inside of the polygon. rove: () The apothems of each polygon are all congruent. (2) The foot of each apothem is on the circle. (3) The sides of the regular polygon are tangent to the circle. c. Let r be the distance from the center to a vertex of the regular polygon. Since the center is equidistant from each vertex, it is possible to circumscribe a circle about the polygon with radius r. Let a be the length of an apothem and s be the length of a side of the regular polygon. How is the radius, r, of the circumscribed circle related to the radius, a, of the inscribed circle? 3-5 NGLS FRM Y TNGNTS, HRS, N SNTS ngles Formed by a Tangent and a hord In the diagram, g is tangent to circle at, is a chord, and is a diameter. When is drawn, is a right angle because it is an angle inscribed in a semicircle, and is the complement of. lso, ', is a right angle, and is the complement of. Therefore, since complements of the same angle are congruent,.we can conclude that since m = X, then m = X 2 m 2 m.

34 568 Geometry of the ircle We can state what we have just proved on page 567 as a theorem. Theorem 3.2 The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. ngles Formed by Two Intersecting hords We can find how the measures of other angles and their intercepted arcs are related. For example, in the diagram, two chords and intersect in the interior of circle and is drawn.ngle is an exterior angle of. Therefore, m m m X 2 m 2 mx X 2 Notice that X (m m X ) is the arc intercepted by and X is the arc intercepted by, the angle vertical to. We can state this relationship as a theorem. Theorem 3.3 The measure of an angle formed by two chords intersecting within a circle is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. ngles Formed by Tangents and Secants We have shown how the measures of angles whose vertices are on the circle or within the circle are related to the measures of their intercepted arcs. Now we want to show how angles formed by two tangents, a tangent and a secant, or two secants, all of which have vertices outside the circle, are related to the measures of the intercepted arcs. Tangent Intersecting a Secant g In the diagram, RS is a tangent to circle at R g and TQ is a secant that intersects the circle at T and at Q. hord RQ is drawn. Then SRQ is an exterior angle of RQ. Q S R T

35 ngles Formed by Tangents, hords, and Secants 569 m RQ m m SRQ m m SRQ m RQ m X X 2 mrq 2 mrt m 2 (mrq X 2 mrt X ) Two Intersecting Secants g In the diagram, TR is a secant to circle g that intersects the circle at R and T, and QS is a secant to circle that intersects the circle at Q and S. hord RQ is drawn. Then RQS is an exterior angle of RQ. m RQ m m RQS R m m RQS m RQ m X X 2 mrs 2 m X mqt 2 (mrs 2 mqt X ) S T Q Two Intersecting Tangents g In the diagram, RS is tangent to circle at R, Q g is tangent to the circle at Q, and T is a point on major RQX. hord RQ is drawn.then SRQ is an exterior angle of RQ. m QR m m SRQ m m SRQ m QR m X X 2 mrtq 2 m X mrq 2 (mrtq 2 mrq X ) S T R Q For each pair of lines, a tangent and a secant, two secants, and two tangents, the steps necessary to prove the following theorem have been given: Theorem 3.4 The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs.

36 570 Geometry of the ircle XML tangent and a secant are drawn to circle from point. The tangent intersects the circle at Q and the secant at R and S. If mqrx : mrs X : msq X 5 3:5:7, find: X a. mqrx b. mrs c. msqx S Q R d. m QRS e. m RQ f. m Solution Let mqrx 3x, mrs 5x, and msqx 7x. X 3x 5x 7x x x 5 24 a. mqr X 5 3x b. mrs X 5 5x c. 5 3(24) 5 5(24) msqx 5 7x 7(24) d. e. m/rq 5 m/qrs 5 X 2 2 mqrx msq f. m/ (68) 5 2 (72) (msq X 2 mqr X ) 5 2 ( ) 5 48 nswers a. 72 b. 20 c. 68 d. 84 e. 36 f. 48 Note: m/qrs 5 m/rq m/ XML 2 Two tangent segments, R and RQ, are drawn to circle from an external point R. If m R is 70, find the measure of the minor arc QX and of the major arc SQ into which the circle is divided. X

37 ngles Formed by Tangents, hords, and Secants 57 Solution The sum of the minor arc and the major arc with the same endpoints is 360. Let x = mqx. X Then 360 x = msq. X 2 mq X ) m R 2 (msq 70 2 (360 2 x 2 x) 70 2 ( x) x x x nswer mqx 0 and msq X 250 S Q R SUMMRY Type of ngle egree Measure xample Formed by a Tangent and a hord Formed by Two Intersecting hords The measure of an angle formed by a tangent and a chord that intersect at the point of tangency is equal to one-half the measure of the intercepted arc. The measure of an angle formed by two intersecting chords is equal to one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. m/ 5 X 2 m 2 m/ 5 2 (m X m X ) m/2 5 2 (m X m X ) (ontinued)

38 572 Geometry of the ircle SUMMRY (ontinued) Type of ngle egree Measure xample Formed by Tangents and Secants The measure of an angle formed by a tangent and a secant, two secants, or two tangents intersecting outside the circle is equal to one-half the difference of the measures of the intercepted arcs. 2 3 m/ 5 2 (m X 2 m X ) X 2 m X ) m/2 5 2 (m m/3 5 X 2 (m 2 m X ) xercises Writing bout Mathematics. Nina said that a radius drawn to the point at which a secant intersects a circle cannot be perpendicular to the secant. o you agree with Nina? xplain why or why not. 2. Two chords intersect at the center of a circle forming four central angles. aron said that the measure of one of these angles is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle. o you agree with aron? xplain why or why not. eveloping Skills g g In 3 8, secants QS and RT intersect at. 3. If mstx 60 and mqrx 90, find m. 4. If mstx 00 and mqrx 40, find m. 5. If mstx 70 and X 0, find m. 6. If m 40 and mqrx 86, find mstx. 7. If m 60 and mqrx 50, find mstx. 8. If m 25 and mstx 0, find mqrx. S Q R T

39 ngles Formed by Tangents, hords, and Secants 573 g In 9 4, tangent Q g and secant RT intersect at. 9. If mqtx 70 and mqrx 70, find m. 0. If mqtx 20 and mqrx 30, find m. R. If mqrx 70 and mrtx 20, find m. Q 2. If mqrx 50 and m 40, find mqtx. 3. If mqrx 60 and m 35, find mqtx. T 4. If m 30 and mqrx 20, find mqtx. In 5 20, tangents R g and Q g intersect at and S is on major arc QRX. 5. If mrqx 60, find m. 6. If mrqx 80, find m. Q X X 7. If mrsq 260, find m. 8. If mrsq 20, find m. X 5 2mRQ X 9. If mrsq, find m. 20. If m 45, find mrqx and mrsq X. R S In 2 26, chords and intersect at in the interior of a circle. 2. If mx 30 and mx 80, find m. 22. If mx 80 and mx 00 find m. 23. If mx 25 and mx 45, find m. 24. If mx 20 and mx 60, find m. 25. If mx 30 and m 50, find mx. 26. If mx 80 and m 30, find mx. 27. In the diagram, g and g are tangent to circle at and. iameter and chord intersect at, mx 25 and m 55. Find: a. mx b. mx c. mx d. m e. m f. m g. Show that is perpendicular to and bisects

40 574 Geometry of the ircle 28. Tangent segment and secant segment are drawn to circle and and are chords. If m 45 and mx : m X 5 5:2, find: a. mx b. mx c. m d. m e. m f. m 45 pplying Skills 29. Tangent g intersects circle at, chord g, diameter intersects at, and diameter F is extended to. a. rove that. b. If m 30, find mx, mfx, mfx, mx, mx, and m. F g 30. Tangent intersects circle at, secant F intersects the circle at F and, and secant G intersects the circle at G and.if mf X 5 mg X, prove that is isosceles. F G 3. Segments and are tangent 32. Secant intersects a circle to circle at and, respectively, at and. hord is drawn. and m 20. rove that rove that m X. is equilateral. 2 m g

41 Measures of Tangent Segments, hords, and Secant Segments MSURS F TNGNT SGMNTS, HRS, N SNT SGMNTS Segments Formed by Two Intersecting hords We have been proving theorems to establish the relationship between the measures of angles of a circle and the measures of the intercepted arcs. Now we will study the measures of tangent segments, secant segments, and chords. To do this, we will use what we know about similar triangles. Theorem 3.5 If two chords intersect within a circle, the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other. Given hords and intersect at in the interior of circle. rove ()() ()() roof Statements Reasons. raw and.. Two points determine a line. 2. and 2. Inscribed angles of a circle that intercept the same arc are congruent The lengths of the corresponding sides of similar triangles are in proportion. 5. ()() ()() 5. In a proportion, the product of the means is equal to the product of the extremes. Segments Formed by a Tangent Intersecting a Secant o similar relationships exist for tangent segments and secant segments? In the diagram, tangent segment is drawn to circle, and secant segment intersects the circle at and.

42 576 Geometry of the ircle We will call, the part of the secant segment that is outside the circle, the external segment of the secant. When chords and are drawn, because the measure of each is one-half the measure of the intercepted arc, X. lso by the reflexive property. Therefore, by. The length of the corresponding sides of similar triangles are in proportion. Therefore: 5 and () 2 ()() We can write what we have just proved as a theorem: Theorem 3.6 If a tangent and a secant are drawn to a circle from an external point, then the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external segment. Note that both means of the proportion are, the length of the tangent segment. Therefore, we can say that the length of the tangent segment is the mean proportional between the lengths of the secant and its external segment. Theorem 3.6 can be stated in another way. Theorem 3.6 If a tangent and a secant are drawn to a circle from an external point, then the length of the tangent segment is the mean proportional between the lengths of the secant segment and its external segment. Segments Formed by Intersecting Secants What is the relationship of the lengths of two secants drawn to a circle from an external F point? Let and be two secant segments drawn to a circle as shown in the diagram. raw F a tangent segment to the circle from. Since F 2 ()() and F 2 ()(), then ()() ()() Note: This relationship could also have been proved by showing that.

43 Measures of Tangent Segments, hords, and Secant Segments 577 We can state this as a theorem: Theorem 3.7 If two secant segments are drawn to a circle from an external point, then the product of the lengths of one secant segment and its external segment is equal to the product of the lengths of the other secant segment and its external segment. XML Two secant segments, and, and a tangent segment,, are drawn to a circle from an external point. If 9 cm, 2 cm, and the external segment of is centimeter longer than the external segment of, find: a. b. c. Solution Let x and x. ()() 5 ()() (x ) 5 2x 9x 9 5 2x 9 5 3x 3 5 x a. x 4 b. x 3 c. () 2 5 ()() () 2 5 (9)(4) () (Use the positive square root.) nswers a. 4 cm b. 3 cm c. 6 cm XML 2 In a circle, chords Q and RS intersect at T. If T 2, TQ 0, and SR 9, find ST and TR. R 2 T 0 Q 9 S