# Lesson 18: Recognizing Equations of Circles

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1 Student Outcomes Students complete the square in order to write the equation of a circle in center-radius form. Students recognize when a quadratic in xx and yy is the equation for a circle. Lesson Notes This lesson builds from the understanding of equations of circles in Lesson 17. The goal is for students to recognize when a quadratic equation in xx and yy is the equation for a circle. The Opening Exercise reminds students of algebraic skills that are needed in this lesson, specifically multiplying binomials, factoring trinomials, and completing the square. Throughout the lesson, students will need to complete the square (A-SSE.b) in order to determine the center and radius of the circle. Completing the square was taught in Grade 9, Module 4, Lessons 11 and 12. Classwork Opening Exercise (6 minutes) It is important that students can factor trinomial squares and complete the square in order to recognize equations of circles. Use the following exercises to determine which students may need remediation of these skills. Opening Exercise xx xx xx 2 xx xx? = 40 xx xx xx 2 xx xx 9 = 49 Date: 10/22/14 227

3 Sometimes equations of circles are presented in this simplified form. To easily identify the center and the radius of the graph of the circle, we sometimes need to factor and/or complete the square in order to rewrite the equation back in its standard form (xx aa) 2 + (yy bb) 2 = rr 2. Exercise 1 ( minutes) Use the exercise below to assess students understanding of the content in Example 1. Exercise 1. Rewrite the following equations in the form (xx aa) 22 + (yy bb) 22 = rr 22. a. xx yy = (xx + 22) 22 + (yy ) 22 = b. xx yy = 44 (xx 55) 22 + (yy + 77) 22 = 44 Example 2 (5 minutes) Scaffolding: Students can draw squares as in the Opening Exercise to complete the square. For advanced learners, offer equations with coefficients on one or both of the squared terms. For example, xx xx + 4yy 2 48yy = 197. Example 2 What is the center and radius of the following circle? xx yy = 4444 Provide time for students to think about this in pairs or small groups. If necessary, guide their thinking by telling students that part (d) in the Opening Exercise is related to the work they will be doing in this example. Allow individual students or groups of students to share their reasoning as to how they determine the radius and center of the circle. MP.1 We can complete the square, twice, in order to rewrite the equation in the necessary form. First, xx 2 + 4xx: xx 2 + 4xx + 4 = (xx + 2) 2. Second, yy 2 12yy: yy 2 12yy + 6 = (yy 6) 2. Then, xx 2 + 4xx + yy 2 12yy = 41 (xx 2 + 4xx + 4) + (yy 2 12yy + 6) = (xx + 2) 2 + (yy 6) 2 = 81. Therefore, the center is at ( 2, 6), and the radius is 9. Exercises 2 4 (6 minutes) Use the exercise below to assess students understanding of the content in Example 2. Date: 10/22/14 229

5 Now rearrange terms using the commutative and associative properties of addition to match xx 2 + yy 2 + AAAA + BBBB + CC = 0. xx 2 + yy 2 2aaaa 2bbbb + (aa 2 + bb 2 rr 2 ) = 0 What expressions are equivalent to AA, BB, and CC? AA = 2aa, BB = 2bb, and CC = aa 2 + bb 2 rr 2, where AA, BB, and CC are constants. So, could xx 2 + yy 2 + AAAA + BBBB + CC = 0 represent a circle? Yes. The graph of the equation xx 2 + yy 2 + AAAA + BBBB + CC = 0 is a circle, a point, or an empty set. We want to be able to determine which of the three possibilities is true for a given equation. Let s begin by completing the square: xx 2 + yy 2 + AAAA + BBBB + CC = 0. xx 2 + yy 2 + AAAA + BBBB + CC = 0 xx 2 + AAAA + yy 2 + BBBB + CC = 0 xx 2 + AAAA + + yy 2 + BBBB + = CC xx 2 + AAAA + AA yy 2 + BBBB + BB 2 2 = CC + AA BB 2 2 xx + AA yy + BB 2 2 = CC + AA BB 2 2 xx + AA yy + BB 2 2 xx + AA yy + BB 2 2 = 4CC 4 + AA2 4 + BB2 4 = AA2 + BB 2 4CC 4 Just as the discriminant is used to determine how many times a graph will intersect the xx-axis, we use the right side of the equation above to determine what the graph of the equation will look like. If the fraction on the right is positive, then we can identify the center of the circle as AA, BB and the radius 2 2 as AA2 +BB 2 4CC 4 = AA2 +BB 2 4CC 2 = 1 2 AA2 + BB 2 4CC. What do you think it means if the fraction on the right is zero? It means we could locate a center, AA, BB, but the radius is 0; therefore, the graph of the equation 2 2 is a point, not a circle. What do you think it means if the fraction on the right is negative? It means that we have a negative length of a radius, which is impossible. When the fraction on the right is negative, we have an empty set. That is, there are no possible solutions to this equation because the sum of two squared numbers cannot be negative. If there are no solutions, then there are no ordered points to graph. We can apply this knowledge to equations whose graphs we cannot easily identify. For example, write the equation xx 2 + yy 2 + 2xx + 4yy + 5 = 0 in the form of (xx aa) 2 + (yy bb) 2 = rr 2. xx 2 + yy 2 + 2xx + 4yy + 5 = 0 (xx + 2xx + 1) + (yy + 4yy + 4) + 5 = 5 (xx + 1) 2 + (yy + 2) 2 = 0 Since the right side of the equation is zero, then the graph of the equation is a point, ( 1, 2). Date: 10/22/14 21

6 Exercise 5 (6 minutes) Use the exercise below to assess students understanding of the content in Example. 5. Identify the graphs of the following equations as a circle, a point, or an empty set. a. xx 22 + yy = 00 xx yy 22 = 44 (xx + 22) 22 + yy 22 = 44 The right side of the equation is positive, so the graph of the equation is a circle. b. xx 22 + yy = 00 (xx ) + (yy ) = 1111 (xx ) 22 + (yy 22) 22 = 22 The right side of the equation is negative, so the graph of this equation cannot be a circle; it is an empty set. c. 22xx yy yy = xx yy yy = xx xx + yy yy = 00 xx xx yy yy = xx yy = xx yy = xx yy = 00 The right side of the equation is equal to zero, so the graph of the equation is a point. Closing (2 minutes) Have students summarize the main points of the lesson in writing, by talking to a partner, or as a whole class discussion. Use the questions below, if necessary. Describe the algebraic skills that were necessary to work with equations of circles that were not of the form (xx aa) 2 + (yy bb) 2 = rr 2. Given the graph of a quadratic equation in xx and yy, how can we recognize it as a circle or something else? Exit Ticket (5 minutes) Date: 10/22/14 22

7 Name Date Exit Ticket 1. The graph of the equation below is a circle. Identify the center and radius of the circle. xx xx + yy 2 8yy 8 = 0 2. Describe the graph of each equation. Explain how you know what the graph will look like. a. xx 2 + 2xx + yy 2 = 1 b. xx 2 + yy 2 = c. xx 2 + yy 2 + 6xx + 6yy = 7 Date: 10/22/14 2

8 Exit Ticket Sample Solutions 1. The graph of the equation below is a circle. Identify the center and radius of the circle. xx yy = 00 (xx + 55) 22 + (yy 44) 22 = 4444 The center is ( 55, 44), and the radius is Describe the graph of each equation. Explain how you know what the graph will look like. a. xx yy 22 = 11 (xx + 11) 22 + yy 22 = 00 The graph of this equation is a point. I can place a center at ( 11, 00), but the radius is equal to zero. Therefore, the graph is just a single point. b. xx 22 + yy 22 = The right side of the equation is negative, so its graph cannot be a circle; it is an empty set. c. xx 22 + yy = 77 (xx + ) 22 + (yy + ) 22 = 2222 The graph of this equation is a circle with center (, ) and radius 55. Since the equation can be put in the form (xx aa) 22 + (yy bb) 22 = rr 22, I know the graph of it is a circle. Problem Set Sample Solutions 1. Identify the center and radii of the following circles. a. (xx ) + yy 22 = 11 The center is ( 2222, 00), and the radius is 11. b. xx yy = 88 (xx + 11) 22 + (yy 44) 22 = 2222 The center is ( 11, 44), and the radius is 55. c. xx yy yy = 00 (xx 1111) 22 + (yy 55) 22 = The center is (1111, 55), and the radius is d. xx 22 + yy 22 = 1111 The center is (00, 00), and the radius is Date: 10/22/14 24

9 e. xx 22 + xx + yy 22 + yy = xx yy = The center is 11 22, 11 22, and the radius is Sketch a graph of the following equations. a. xx 22 + yy = 00 (xx + 55) 22 + (yy 22) 22 = 44 The right side of the equation is negative, so its graph cannot be a circle; it is an empty set, and since there are no solutions, there is nothing to graph. b. xx 22 + yy = 00 (xx + 77) 22 + (yy 88) 22 = 99 The graph of the equation is a circle with center ( 77, 88) and radius. Date: 10/22/14 25

10 c. xx 22 + yy = 00 (xx + 22) 22 + (yy 55) 22 = 00 The graph of this equation is a point at ( 22, 55).. Chante claims that two circles given by (xx + 22) 22 + (yy 44) 22 = 4444 and xx 22 + yy = 00 are externally tangent. She is right. Show that she is. (xx + 22) 22 + (yy 44) 22 = 4444 This graph of this equation is a circle with center ( 22, 44) and radius 77. (xx ) 22 + (yy + 88) 22 = This graph of this equation is a circle with center (, 88) and radius 66. By the distance formula, the distance between the two centers is 1111, which is precisely the sum of the radii of the two circles. Therefore, these circles will touch only at one point. Put differently, they are externally tangent. 4. Draw a circle. Randomly select a point in the interior of the circle; label the point AA. Construct the greatest radius circle with center AA that lies within the circular region defined by the original circle. Hint: Draw a line through the center, the circle, and point AA. Constructions will vary. Ensure that students have drawn two circles such that the circle with center AA is in the interior and tangent to the original circle. Date: 10/22/14 26

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