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Kapitolyo High School

Learn how to identify tangent and secant lines, graph linear and circular equations, and solve for the point of intersection between a circle and a tangent or secant line. Prepare for an exploration of circles!

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Kapitolyo High School

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  1. Kapitolyo High School S I M Strategic Intervention Materials in GEOMETRY Mathematics III

  2. Prepared by: and Rod Umali Leonelyn P. Dela Cruz

  3. Belen A. Bahia Master Teacher I/Dep’t Coordinator Domingo N. Viñas Principal II

  4. Objectives : At the end of the session , the selected students will be able to: • Identify the difference between a tangent line and a secant line ; • 2. Illustrate the linear and circular equations by graphing ; • 3. Solve systematically and neatly the point of intersection between a circle and a tangent line and between a circle and a secant line.

  5. Let’s discover the amazing world of CIRCLES…

  6. CIRCLES IN THE COORDINATE PLANE TITLE CARD

  7. … I remember how to find the equation of a circle, given the radius and the center…but what if there are lines intersecting the circles? Oh, it looks like a radar in a plane… isn’t it? Yes, but it has squares on it... Yes, you’re right ! But I am just curious on how to find the point of intersection between a line and a circle… Our teacher told us that if a line intersects a circle at one point, the line is called a TANGENT LINE.. …and if a line intersects a circle at two distinct points, the line is called a SECANT LINE.

  8. It’s better if we ask our teacher how to find the point of intersection between the circle and a tangent line…and between a circle and a secant line… let’s go!!! A line may intersect a circle at two points, at one point or at no point at all… GUIDE CARD

  9. But before we proceed , you must know first , how to identify the equation of the circle from the equation of the line. #1 is an equation of a line with y-intercept of 9 and a slope of -1 1.) x + y = 9 equation of the line 2.) x2 + y2 = 100 3.)( x + 4 )2 + ( y – 2 )2 = 52 Equations of the circle #3 is an equation of a circle with center at (-4 , 2 ) and a radius of 5 # 2 is an equation of a circle with center at the origin and a radius of 10

  10. To get the product, Square the 1st term, multiply the product of the 1st term and 2nd term by 2, then square the 2nd term. You will also use your skills in Algebra to expand the square of a binomial… Ex. (y – 3 )2 = y2 – 6y + 9

  11. …and develop your skills on how to get the square root of a number ,how to get the product of a given number raised to a certain exponent, how to square a binomial ,how to get the factors of the perfect square trinomial or any given polynomials … Oh, I have to review my algebra … …too many…but I know I can do it…

  12. …and you must also know how to identify the radius and the center given the equation of the circle and vice versa … I know you can do it by pair…ok? Yes, Sir!!!

  13. In the equation ( x – 6 )2 + ( y + 4 )2 = 81, what is the center? x – 6 = 0 y + 4 = 0 x = 6 y = - 4 The center is ( 6 , - 4 ) How about the radius of that equation? Easy! Just get the square root of 81…The radius is 9…

  14. Yes, I do… the answers are -11 and +11… 121… Do you know the square root of 121 ? How did you do it? Just think of a number in which when you multiply the number by itself twice, the product is 121…

  15. 54 … just multiply the base which is 5 , four times by itself…4 which is the exponent , tells us how many times we multiply the base by itself… so, 54 = 5x5x5x5 = 25x25 = 625… She’s studious and diligent in the class, I will ask her… Hi, Ms. Butterfly, I’m Spider, your classmate in Geometry… Can you help me find the value of 54? Thank you… You’re welcome…

  16. Get the square root of the 1st term, then think of the factors of the 3rd term, such that when you add them, the result is the numerical coefficient of the 2nd term… Penguin, do you know the factors of t2 +2t - 35? The answer is ( t +7 ) ( t - 5 ) Are you sure with your answer? Yes, I’m sure with my answer…

  17. ( 3x – 4 )2 = ( 3x – 4 ) ( 3x – 4 ) = 9x2 – 12x – 12x + 16 ( 3x – 4 )2 = 9x2 – 24x + 16 Can you help me find the product of ( 3x – 4 )2? Ok, just square the 1st term, multiply the product of the 1st & 2nd term by 2, then square the second term… or use the happy face method… Ok, thanks… I’ll practice it at home... You’re welcome…

  18. Earlier, you ask me how to solve for the point of intersection of the circle and a tangent line. Here, I will give you two equations like x2 + y2 = 20 x – 2y = 10 We have to use eq. 2 to solve for x… eq.1 eq. 2 x2 + y 2= 20 x – 2y = 10 x – 2y= 10 x = 2y+ 10 eq.1 eq. 2 So, the value of x is 2y + 10

  19. Simplify(2y + 10 )2, it becomes 4y2 + 40y + 100. …to solve for y, substitute the value of X in eq. 1…. …then add similar terms… X = 2y + 10 x2 + y2 = 20 (2y + 10)2 + y2 = 20 4y 2+ 40y + 100 + y2 = 20 5y2+ 40y + 80 = 0 5 ( y2 + 8y + 16 ) = 0 5 (y + 4 )2 = 0 y = -4 …take out the common factor… …so, the value of y = -4 …get the factors of (y2 + 8y + 16 )… we have ( y + 4 )2…

  20. To solve for x, substitute the value of yin eq. 2 y = -4 x – 2y = 10 x– 2( -4 ) = 10 x + 8 = 10 x = 10-8 x= 2 ss: ( 2 , -4) ( x , y ) …the solution set is ( 2 ,- 4 ).

  21. …here is the graph of the equations: x2 + y2 = 20 andx – 2y = 10…Thecircle and the line intersect at point ( 2 , - 4)… ...Do you have any clarifications?... If none, let’s proceed to example # 2… Y 4 x2 + y2 = 20 3 2 1 X -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -1 -2 -3 x– 2y = 10 -4 ( 2, - 4 ) -5

  22. Our next example will teach you how to solve for the point of intersection of the circle and a secant line, here are the equations: x2 + y2 = 25 x – y + 1 = 0 We will solve for xusing eq.2 … eq.1 eq. 2 x2 + y 2= 25 x – y + 1 = 0 x – y + 1 = 0 x = y - 1 eq.1 eq. 2 So, the value of x is y - 1

  23. …simplify ( y - 1 )2, it becomes y2 - 2y +1… …substitute the value of X in eq. 1… x = y -1 x2 + y2 = 25 (y - 1)2 + y2 = 25 y 2- 2y + 1 + y2 = 25 2y2- 2y -24= 0 2( y2 - y - 12) = 0 2 ( y + 3 ) (y - 4 )= 0 y = -3 y = 4 …then add similar terms… …take out the common factor… …get the factors of (y2 - y - 12)…these are: ( y + 3 ) and ( y – 4 )

  24. …there are two values of y… we have , y = -3 and y = 4…substitute them to eq. 2… If y = -3 If y = 4 x = y – 1 x = y - 1 x= -3 – 1 x = 4 - 1 x = -4x = 3 ss: ( -4, -3) ss: ( 3 , 4) ( x , y ) ( x , y ) …It means that the intersection of the circle x2 + y2 = 25 and a line x – y + 1 = 0 is at (-4,-3) and (3,4)… That means, the line is a secant line because it touches the circle at two points…

  25. …any clarifications with Ex. # 2 ?... …If none, here’s the graph of x2 + y2 = 25 andx –y + 1 = 0…The circle and the line intersect at point ( - 4 ,- 3 ) and ( 3,4 )… Y x2 + y2 = 25 5 ( 3,4 ) 4 3 2 1 X -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 (-4 , -3) -3 -4 -5 x– y + 1 = 0

  26. ACTIVITY CARD

  27. Activity 1 RIDDLE When the clock strikes 13, what time is it? To find the answer to this riddle , identify the center and the radius of each equation, then use the decoder to reveal the answer. 1.) x2 + y2 = 121 2.) ( x – 4 )2 + y2 = 16 3.) x2 + ( y – 3 )2 = 10 TIME C ( 0 , 0 ) , r =11 4.) x2 + y2 + 4x – 6y = - 4 5.) x2 + y2 – 10x – 6y = - 18 GETC ( 0 , 3 ) r= 10 IT C ( -2, 3 ) , r = 3 TO C ( 4 , 0 ) , r = 4 FIXED C ( 5 , 3 ) r = 4 DECODER

  28. Activity 2 MAZE What did the circle say to the tangent line? You can answer this by helping the four mice to the right path in a maze and collect all the letters that corresponds to the equations inside the maze and decode it to the boxes below. • 1. C ( 1 , 2 ) , r = 4 5. C ( -5 , 2 ) , r = 10 • 2 . C ( 0 , 0 ) , r = 8 6. C ( 4 , -2 ) , r = 7 • 3. C ( 4 , -7 ) , r = 3 7. C ( 0 , 0 ) , r = 2 2 • 4. C ( -2 , -5 ) , r = 11

  29. MAZE ( x - 1 )2 + ( y – 2 )2 = 16 x 2 + y2 = 64 x 2 + y2 - 8x + 14y = -56 x 2 + 4x + y 2 + 10y = 92 x2 + 10x + y 2 - 4y = 71 x 2 + y 2- 8x + 4y = 29 x 2 + y 2 = 8 UCx 2 + 4x + y 2 + 10y = 92 ST ( x - 1 )2 + ( y – 2 )2 = 16 NGx 2 + y 2- 8x + 4y = 29 HI x2 + 10x + y 2 - 4y = 71 OPx 2 + y2 = 64 TOx 2 + y2 - 8x + 14y = -56 ME x 2 + y 2 = 8 1 2 3 4 5 6 7

  30. Find the intersection of a line and a circle. 1.) x2 + y2 = 9 x + y = 3 2.) x2 + y2 = 16 x - y = 4 ASSESSMENT CARD # 1

  31. (1 ) ( 2 ) ( 5 ) ASSESSMENT CARD # 2 ( 4 ) ( 3 )

  32. Student’s Note Please check the circles below: After reading this lesson, I now understand the following concepts: Tangent line intersects the circle at exactly one point; Secant line intersects the circle at two distinct points; How to find the point of intersection between a circle and a tangent line; How to find the point of intersection between a circle and a secant line; Finding the radius and the center given the equation of the circle in which the center is found at the origin; Finding the radius and the center given the equation of the circle in which the center is not found at the origin; Finding the equation of the circle given the radius and the center; Writing the general form of the equation of the circle in center-radius form.

  33. Student’s Note The activities have been: Easy Average Difficult Now, I can do the activities : Alone With one of my classmates to work with Within a group

  34. ENRICHMENT CARD

  35. Theater Lighting • A bank of lights is arranged over a stage. Each light illuminates a circular area on the stage. A coordinate plane is used to arrange the lights, using the corner of the stage as the origin . The equation ( x – 13 )2 + ( y – 4 )2 = 16 represents one of the disks of light. • Graph the disk of light. • Three actors are located as follows : Leonel is at ( 11 , 4 ), • Rod is at ( 8 , 5 ) and Philip is at ( 15 , 5 ) . Which actors are in the disk of light? Enrichment 1 Enrichment 1

  36. CELL PHONES A cellular phone network uses towers to transmit calls. Each towers transmits to a circular area. On a grid of a city, the coordinates of the location and the radius each tower covers are as follows ( integers represent kilometers): Tower A is at ( 0 , 0 ) and covers a 3 km radius, Tower B is at ( 5 , 3 ) and covers a 2.5 km radius, and Tower C is at ( 2 , 5 ) and covers a 2 km radius. a. Write the equation s that represent the transmission boundaries of the two towers . Graph each equation. b. Tell which towers , if any, transmit to a phone located at J ( 1 , 1 ) , K ( 4 , 2 ), L ( 3.5 , 4.5 ) M ( 2 , 2.8 ) , or N ( 1 , 6 ). Enrichment 2 Enrichment 2

  37. THE ANSWER KEY CARD

  38. Answer for Activity Card # 1 Answer for Activity Card # 2 S T O P T O U C H I N G M E 1 2 3 4 5 6 7

  39. Assessment Card # 1 Assessment Card # 2 1 1 ( -5 , -2 ) and ( 2 , 5 ) 1 1 ( 3 , 0 ) and ( 0 , 3 ) 2 2 (-12,-5 ) and ( 5 , 12 ) 2 2 ( 4 , 0 ) and ( 0 , -4 ) 3 3 ( 0, 5 ) 3 3 ( 4 , 3 ) and (-4 , 3 ) 4 ( 0 , 20 ) and ( 20 , 0 ) 4 ( 6 , 8 ) and ( 6 , -8 ) 5 ( 0 , 15 ) and ( 15 , 0 )

  40. Answers for Enrichment Card # 1 • Rewrite the equation to find the center and radius: • ( x – 13 )2 + ( y – 4 )2 = 16 • ( x – 13 )2 + ( y – 4 )2 = 42 • The center is ( 13, 4 ) and the radius is 4. The circle is shown below. Y 6 Philip ( 15 , 5 ) Leonel ( 11 , 4 ) Rod ( 8 , 5 ) 4 ( 13 , 4 ) 2 X 5 10 15 b. The graph shows that Leonel and Philip are both in the disk of light.

  41. Answers for Enrichment Card # 2 • Tower A ( x2 + y)2 = 9 • Tower B ( x – 5 )2 + ( y – 3 )2 = 6.25 • Tower C ( x – 2 )2 + ( y – 5 )2 = 4 Y 6 N C b. J ( 1, 1 ) – Tower A K ( 4 , 2 ) – Tower B L ( 3.5 , 4.5 ) – Tower B & C M ( 2 , 2.8 ) – No signal N ( 1 , 6 ) – Tower C 5 L 4 B 3 M 2 K 1 J A X 0 1 2 3 4 5

  42. BOOKS Geometry, Revised Edition (NPSBE) Dilao , Soledad J. et al, SD Publications , Inc. Quezon City, 2009 Geometry ( Applying , Reasoning, Measuring ) Larson, Ron et al, McDougal Littell , Illinois, 2004 Work Text in Geometry Simplified Concepts and Structures Pascual, Ferdinand C. et al , Innovative Educational Materials , Inc, Manila, 2002 Geometry, Teacher’s Edition Clemens , Stanley R. et al, Addison-Wesley Publishing Company, Inc. , USA, 1994 SOFTWARE Dynamic Geometry Software for Exploring Mathematics Version 4.06, KCP Technology, Inc. 2001 WEBSITE http://www.classzone.com http://www.mcdougalittell.com http://www.keypress.com/sketchpad R E F E R E N C E C A R D

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