Questions and Answers #3 Circles

Questions and Answers Sheet 3 Circles Question #1 Two circles have equations of 𝑥 2 + 𝑦2 − 4𝑥 − 4𝑦 + 4 = 0 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 + 4 = 0 Find the distance between the centers of the two circles. Answer: Step 1 Equations of two circles are: 𝐶1 : 𝑥 2 + 𝑦 2 − 4𝑥 − 4𝑦 + 4 = 0 𝐶2 : 𝑥 2 + 𝑦 2 − 4𝑥 + 8𝑦 + 4 = 0 Converting these equations in theria standard form 𝐶1 : 𝑥 2 + 𝑦 2 − 4𝑥 − 4𝑦 + 4 = 0 𝑥 2 − 4𝑥 + 4 + 𝑦 2 − 4𝑦 + 4 = 4 1) 𝐶1 : (𝑥 − 2)2 + (𝑦 − 2)2 = (2)2 𝐶2 : 𝑥 2 + 𝑦 2 − 4𝑥 − 8𝑦 + 4 = 0 𝑥 2 − 4𝑥 + 4 + 𝑦 2 − 8𝑦 + 16 = 16 2) 𝐶2 : (𝑥 − 2)2 + (𝑦 − 4)2 = (4)2 Step 2 In 𝐶1 , Centers is (2, 2) and radius = 2 In 𝐶2 , Centers is (2, 4) and radius = 4 Therefore distance 𝑑 = √(4 − 𝑑=2 2)2 𝑏 𝑤 the Centers is given by: + (2 − 2) = √4 = 2 Question #2 In the given figure, X and Y are centers of two intersecting equal circle. If 𝑋𝑌 = 𝐴𝐵 , prove that AXBY is a square. Answer: Step 1 It is given that X and Y are the centers of two intersecting equal circles by which we can say that the radius of both circles is equal. Two circles are equal if they have the same radius. Also, it is given that 𝑋𝑌 = 𝐴𝐵 To Prove: 𝐴𝑋𝐵𝑌 is a Square Step 2 Proof Since the radius are equal Therefore we can say that 𝐴𝑋 = 𝐴𝑌 = 𝐵𝑌 = 𝑋𝐵 Since all the sides are equal to a quadrilateral and also the diagonals 𝑋𝑌 = 𝐴𝐵 Again, In △ 𝐴𝑋𝐵 and △ 𝐴𝑋 = 𝐴𝑌 (radius of same circle) 𝐵𝑋 = 𝐵𝑌 (radius of same circle) 𝑋𝑌 = 𝐴𝐵 (given) Therefore, By SSS (Side-Side-Side) congruence rule △ 𝐴𝑋𝐵 ≅△ 𝐴𝑌𝐵 By CPCT ∠𝐴𝑋𝐵 = ∠𝐴𝑌𝐵 We know the sum of the opposite angles of a quadrilateral is, Therefore, we have, ∠𝐴𝑋𝐵 + ∠𝐴𝑌𝐵 = 180∘ 2∠𝐴𝑋𝐵 = 180∘ ∠𝐴𝑋𝐵 = 90∘ Similarly, ∠𝐴𝑋𝐵 = ∠𝑋𝐵𝑌 = ∠𝐵𝑌𝐴 = ∠𝑋𝐴𝑌 = 90∘ Step 3 All the sides of the quadrilateral 𝐴𝑋𝐵𝑌 are equal. Diagonals of the quadrilateral 𝐴𝑋𝐵𝑌 are equal. All angles of the quadrilateral 𝐴𝑋𝐵𝑌 are 90∘ Hence 𝐴𝑋𝐵𝑌 is a square Hence Proved. Question #3 Find all the intersection points of the following two circles: - The center circle (0,0) And its radius √13 - The center circle (1, −1) And its radius √17 (Note: Write the answer as follows (𝑥1 , 𝑦1 )(𝑥2 , 𝑦2 )) Answer: Let the equation of circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 Where, h and k are the x and y coordinates of the center of the circle. And r is the radius of the circle. Find the equation of circles by given information, 2 (𝑥 − 0)2 + (𝑦 − 0)2 = (√13) 𝑥 2 + 𝑦 2 = 13 𝑥 2 + 𝑦 2 − 13 = 0 (1) 2 2 (𝑥 − 1)2 + (𝑦 − (−1)) = (√17) (𝑥 − 1)2 + (𝑦 + 1)2 = 17 𝑥 2 + 1 − 2𝑥 + 𝑦 2 + 1 + 2𝑦 = 17 𝑥 2 + 1 − 2𝑥 + 𝑦 2 + 1 + 2𝑦 − 17 = 0 (2) Multiply equation (1) by -1 and add both the equation: −𝑥 2 − 𝑦 2 + 13 = 0 After adding: 1 − 2𝑥 + 1 + 2𝑦 − 4 = 0 −2𝑥 + 2𝑦 − 2 = 0 −𝑥 + 𝑦 − 1 = 0 𝑦 =𝑥+1 Substitute the value of y in equation (1), 𝑥 2 + 𝑦 2 − 13 = 0 𝑥 2 + (𝑥 + 1)2 − 13 = 0 𝑥 2 + 𝑥 2 + 1 + 2𝑥 − 13 = 0 2𝑥 2 + 2𝑥 − 12 = 0 𝑥2 + 𝑥 − 6 = 0 𝑥 2 + 3𝑥 − 2𝑥 − 6 = 0 𝑥 2 + 3𝑥 − 2𝑥 − 6 = 0 𝑥(𝑥 + 3) − 2(𝑥 + 3) = 0 (𝑥 + 3)(𝑥 − 2) = 0 𝑥 = −3,2 𝑥1 = −3, 𝑥2 = 2 𝑦1 = −3 + 1 𝑦1 = −2 𝑦2 = 2 + 1 𝑦2 = 3 𝑦1 = −2, 𝑦2 = 3 SNK Intersection points (𝑥1 , 𝑦1 ) and(𝑥2 , 𝑦2 )are(−3, −2)and(2,3) . Question #4 Find the perpendicular orbits of the family of circles whose center is on the line 𝑦 = 𝑥 and passes through its origin. Answer: Given: the family of circles whose center is on the line 𝑦 = 𝑥 and passes through its origin. To find: The differential equation of the family of circles whose center is on the line 𝑦 = 𝑥 and passes through its origin. Concept Used: The equation of the circle having center as (ℎ, 𝑘) and radius a is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑎2 , using this and substituting appropiate the value of ℎ, 𝑘 and solving accordingly. Explanation Let the center of the circle is (ℎ, 𝑘) and radius is a, as per the question, circles whose center is on the line 𝑦 = 𝑥 and passes through its origin. So, we can write as ℎ = 𝑘 and 𝑎2 = ℎ2 + 𝑘 2 . Now, from the equation of circle can be written as, (𝑥 − ℎ)2 + (𝑦 − ℎ)2 = ℎ2 + 𝑘 2 𝑥 2 + ℎ2 − 2𝑥ℎ + 𝑦 2 + ℎ2 − 2𝑦ℎ = ℎ2 + ℎ2 𝑥 2 + 𝑦 2 − 2𝑥ℎ − 2𝑦ℎ = 0 Now, differentiate the above expression w.r.t. x, we get, 𝑑𝑦 𝑑𝑦 2𝑥 + 2𝑦 ⋅ − 2 ⋅ 1 ⋅ ℎ − 2 ⋅ ℎ ⋅ = 0 𝑑𝑥 𝑑𝑥 𝑑𝑦 (2𝑦 − 2ℎ) − 2ℎ + 2𝑥 = 0 𝑑𝑥 Answer: The differential equation of the family of circles whose center is on the line 𝑦 = 𝑥 and passes 𝑑𝑦 (2𝑦 − 2ℎ) − 2ℎ + 2𝑥 = 0 . through its origin is 𝑑𝑥 Question #5 Find the area for the following circles. Leave your answers in terms of 𝜋 . Answer: 1) Given radius, 𝑟 = 3 Area 𝐴 = 𝜋𝑟 2 = 𝜋(3)2 = 9𝜋 2. diameter, 𝑑 = 12 𝑟= 𝑑 2 = 12 2 =6 2 Area = 𝜋𝑟 = 𝜋62 = 36𝜋 3. diameter, 𝑑 = 8 𝑟= 𝑑 2 8 = =4 2 Area = 𝜋𝑟 2 = 𝜋42 = 16 Question #6 Select from the following two-variable quadratic equations those which are equations of circles. Determine the center and the radius of the circles. a. 𝑥 2 + 𝑦 2 + 6𝑥 − 4𝑦 + 4 = 0 b. 𝑥 2 + 𝑦 2 + 2𝑥𝑦 + 5𝑥 − 3𝑦 − 5 = 0 c. 𝑥 2 + 𝑦 2 − 10𝑥 + 6𝑦 + 35 = 0 d. −2𝑥 2 − 2𝑦 2 + 4𝑥 + 8𝑦 + 22 = 0 Answer: The general equation of circle is 𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 center is (−𝑔, −𝑓) and radius is √82 + 𝑓 2 − 𝑐 a) 𝑥 2 + 𝑦 2 + 6𝑥 − 4𝑦 + 4 = 0 Comparing this equation with the general equation we get: 2𝑔 = 6 ⇒ 𝑔 = 3 2𝑓 = −4 ⇒ 𝑓 = −2 𝑅 = √32 + (−2)2 − 4 = 3 Therefore the center is (−3,2) and radius is 3. b) 𝑥 2 + 𝑦 2 + 2𝑥𝑦 + 5𝑥 − 3𝑦 − 5 = 0 This is not an equation of circle c) 𝑥 2 + 𝑦 2 − 10𝑥 + 6𝑦 + 35 = 0 This can be written as (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 (𝑥 − 5)2 + (𝑦 + 3)2 = 5.572 Therefore the center is (5, −3) and radius is 5.57. d) −2𝑥 2 − 2𝑦 2 + 4𝑥 + 𝑦 + 22 = 0 Divide the equation by -2 𝑥 2 + 𝑦 2 − 2𝑥 − 4𝑦 − 11 = 0 From comparing the above equation with the general equation we get: 2𝑔 = −2 ⇒ 𝑔 = −1 2𝑓 = −4 ⇒ 𝑓 = −2 𝑅 = √12 + 22 + 11 = √1 + 4 + 11 = 4 Therefore, center is (1,2) and radius is 4. Question #7 Find the perimeter for the following figures. All arcs shown are the arcs of circles. Either leave in terms of 𝜋 by entering answers with the word pi, or use 3.14 for pi and round your answer to the nearest 10th. Answer: 1) There are 2 semi circles and one rectangle in the given figure. So, from the given figure we can say, total perimeter = 2(perimeter of one semi circle) + 2(length of rectangle) perimeter of semi circle = 𝜋𝑟 where, 𝑟 = radius of circle = 2 perimeter of semi circle = 2 𝜋 units Total perimeter = 2(perimeter of semi circle) + 2(length of rectangle) = 4𝜋 + 16 units 2) We can see there is a square with corners cut in circular shape. So, total perimeter = (perimeter of inner square of side length 10) + 4(arc length of corner circles) = (perimeter of inner square of side length 10) + 4(circumference of quarter circle) 1 𝜋𝑟 circumference of quarter circle = (2𝜋𝑟) = units 4 2 perimeter of square = 4 (side length) units side of square = 10 units 16−10 radius of circle = = 3 units 2 𝜋𝑟 total perimeter = 4(side) + 4 ( ) 2 3𝜋 = 4(10) + 4 ( ) 2 = 40 + 6 𝜋 units 1) 4𝜋 + 16 2) 6𝜋 + 40 Question #8 Find the equation of a circle whose center is at (4, 1) and tangent to the circle 𝑥 2 + 𝑦 2 − 8𝑥 − 6𝑦 + 9 = 0 . Draw the circles. Answer: We have to find the equation of circle having a center (4,1) and tangent to the circle 𝑥 2 + 𝑦 2 − 8𝑥 − 6𝑦 + 9=0 The equation of the given circle is 𝑥 2 + 𝑦 2 − 8𝑥 − 6𝑦 + 9 = 0 ⇒ 𝑥 2 − 8𝑥 + 16 − 16 + 𝑦 2 − 6𝑦 + 9 − 9 + 9 = 0 ⇒ (𝑥 − 4)2 + (𝑦 − 3)2 − 16 = 0 ⇒ (𝑥 − 4)2 + (𝑦 − 3)2 = 16 ⇒ (𝑥 − 4)2 + (𝑦 − 3)2 = 42 center (4,3) and radius 𝑅 = 4 Now we have to circles having centers 𝐶1 (4,1) and 𝐶2 (4,3) both circles touch each other, then 𝐶1 𝐶2 = √(4 − 4)2 + (3 − 1)2 𝐶1 𝐶2 = √(0)2 + (2)2 𝐶1 𝐶2 = √4 𝐶1 𝐶2 = 2 Since 𝐶1 𝐶2 < 𝑅 so the point (4,1) inside the circle. let radius of circle having center 𝐶1 is r. then 𝑅 − 𝑟 = 𝐶1 𝐶2 4−𝑟=2 −𝑟 = 2 − 4 −𝑟 = −2 𝑟=2 equation of circle having a center (4,1) and radius 𝑟 = 2 is (𝑥 − 4)2 + (𝑦 − 1)2 = 22 𝑥 2 − 8𝑥 + 16 + 𝑦 2 − 2𝑦 + 1 = 4 𝑥 2 + 𝑦 2 − 8𝑥 − 2𝑦 + 17 = 4 𝑥 2 + 𝑦 2 − 8𝑥 − 2𝑦 = 4 − 17 𝑥 2 + 𝑦 2 − 8𝑥 − 2𝑦 + 13 = 0 (required equation of circle) Question #9 Find the area of the shaded region (the circles are edge with all sides of the rectangle). a) 31.6𝑐𝑚2 b) 173.4𝑐𝑚2 c) 226.2𝑐𝑚2 d) 89.0𝑐𝑚2 e) 288𝑐𝑚2 Answer: The correct answer is option A. The area of the given rectangle is 𝐴𝑟 = 𝑙 × 𝑏 = 21 × 7 = 147𝑐𝑚2 7 The radius of the circles will be 𝑟 = 𝑐𝑚 2 The area of a circle will be 𝐴 = 𝜋𝑟 2 7 2 = 𝜋( ) 2 = 38.48𝑐𝑚2 So the area of the three circles will be 𝐴𝐶 = 115.45𝑐𝑚2 So the area of shaded region is the difference between the area of rectangle and area of circles, 𝐴 = 𝐴𝑅 − 𝐴𝐶 = 147 − 115.45 = 31.6𝑐𝑚2 Therefore the correct answer is option A. Question #10 Find the similarity ratio of two circles with areas 75𝜋𝑐𝑚2 and 27𝜋𝑐𝑚2 . Answer: Area of the first circle, 𝐴1 = 75𝜋𝑐𝑚2 Area of the second circle, 𝐴2 = 27𝜋𝑐𝑚2 Let the radius of the circles be 𝑟1 and 𝑟2 respectively. 𝐴1 : 𝐴2 = 75𝜋: 27𝜋 pi cancels pi and we have both 27 and 75 divisible by 3 So, we can simplify this as 𝐴1 : 𝐴2 = 25: 9 ….(1) The scale (similarity) factor for circles is the ratio of the radius Similarity factor = 𝑟1 : 𝑟2 Area of a circle = 𝜋𝑟 2 so 𝐴1 : 𝐴2 = 𝜋𝑟12 : 𝜋𝑟22 This simplifies to 𝐴1 : 𝐴2 = 𝑟12 : 𝑟22 … ..(2) From 1 and 2, we have 𝑟12 : 𝑟22 = 25: 9 Taking square root both sides we get 𝑟1 : 𝑟2 = 5: 3 So, the similarity ratio is 5: 3 [Answer] Question #11 What are the areas of the circles? Use 3.14 for 𝜋 Answer: Step 1 Are of circle = 𝜋 × 𝑟 2 a) Where is the radius of circle So, 𝑟 = 5𝑓𝑡 Area = 𝜋𝑟 2 = 3.14 × (5)2 = 3.14 × 25 Answer: = 78.5𝑓𝑡 2 Question #12 What are the circumference of the circles? Use 3.14 for 𝜋 a) b) Answer: Step 1 a) Given: Diameter = 16 𝑦𝑑 Therefore, diameter = 2𝑅 Circumferense = 2𝜋𝑅 R - radius of circle 16 Circumference = 2𝜋 × 2 Circumference = 16𝜋 = 16 × 3.14 Circumference = 50.24 𝑦𝑑 Question #13 The radil of circles B,K, and D are 6 feet. 14 feet, and 3 feet, respèclivery Part A: Determnine the length of 𝐼𝐾 . Justify your answer. Part B: Determine the lengtn of 𝐸𝐾 . Justify your answer. Part C: Determine the lerath of 𝐵𝐸 . Jusify your dniwer Answer: Step 1 Given: The radii of circles B,K and D are 6 feet, 14 feet and 3 feet respectively. a) length of 𝐼𝐾 = radius of circle - radius of circle B = 14 − 6 = 8 length of 𝐼𝐾 = 8 feet b) length of 𝐸𝐾 = radius of circle K - radius of circle D = 14 − 3 = 11 length of 𝐸𝐾 = 11 feet c) length of 𝐵𝐸 = diameter of circle K - radius of circle B = 2(14) − 3 = 25 length of 𝐵𝐸 = 25 feet Question #14 A circle has a center at (1, 2) and passes through (4, 7) . What is the length of an arc covering \frac{\pi}{4} radians on the circle? Answer: Step 1 Let us denote by S the Circle, by r radius, and pt. 𝐶(1, 2) = Centre of S, and a pt 𝑃(4, 7) on S Then, dist 𝐶𝑃 = 𝑟 , and, using Dist. Formula, we get, 𝑟 2 = 𝐶𝑃2 = (4 − 1)2 + (7 − 2)2 = 9 + 25 = 34 giving 𝑟 = √34 The arc subtends angle 𝜃= \𝑝𝐼 4 at the centre, so that, Length of Arc 𝜋 3.41 = 𝑟 × 𝜃 = √34 × ≅ 5.84 × = 4.58 unit 4 4 Question #15 A circle has a chord that goes from is the length of the chord? Answer: Step 1 𝜋 𝜋 A chord that goes from 2 to 5 3 8 so it travels the distance 𝜋 𝜋 𝜋 2 −5 = 3 8 24 or 𝜋 24 ÷ 2𝜋 = 1 48 of the Circumference of the Circle Area of the Circle = 𝜋𝑟 2 = 120 or 𝑟 2 = 120 or 𝑟 = √120 or 𝑟 = 10.95 Circumference of the circle = 2𝜋𝑟 = 2(𝜋)(10.95) = 68.83 Therefore Length of the chord 1 = 68.83 ( ) = 1.43 48 2𝜋 3 to 5𝜋 8 radians on the circle. If the area of the circle is 120𝜋 , what Question #16 Points (5,4) and (2,2) are 5𝜋 4 radians apart on a circle. What is the shortest arc length between the points? Answer: Step 1 Given: (5, 4) is a point on the circle (2, 2) is a point on the circle Angle subtended at the center is 5𝜋 4 which is reflex lying on the major arc The minor arc happens to be 5𝜋 3𝜋 2𝜋 − = 4 4 The triangle formed by the two points and the center happens to be an is oceles triangle with the shortest angle formed at the center being 3𝜋 𝛼= 4 Remaining angle is (𝜋 − 3𝜋 4 )= 𝜋 4 Angle at each of the vertex at the base is 1𝜋 𝜋 𝜃= = 24 8 Mid point happens to be ( 5+2 2 , 4+2 2 ) = (3.5, 3) We calculate the radius of circle by pythagoras theorem in the right angled triangle formed by one of the point, say (5, 4), the mid-point (3.5, 3), and the center of the circle. Distance between one of the points, say (5, 4) and the mid point (3.5, 3) is given by 𝑎2 = (3.5 − 5)2 + (3 − 4)2 = (−1.5)2 + (−1)2 = 2.25 + 1 = 3.25 𝑎2 = 3.25 Also, the line joining (5, 4) and (3.5, 3) happens to be the adjacent side and the radius being hypotenuse for the angle at the point (5, 4) Consideing the ratio cos 𝜃 𝑎2 cos 2 𝜃 = 2 𝑟 𝜋 𝜃= 8 𝑎2 = 3.25 𝜋 3.25 cos 2 ( ) = 2 8 𝑟 cos2 (𝜋) = 0.146 8 0.854 = 𝑟2 = 3.25 3.25 0.854 𝑟2 = 3.808 𝑟 = √3.808 𝑟 = 1.951 Hence, the shortest arc happens to be 3𝜋 𝑠 = 𝑟𝛼 = 1.951 4 𝑠 = 1.533 the shortest arc happens to be 𝑠 = 4.957 Question #17 A circle has a center at (7, 6) and passes through (2, 1). What is the length of an arc covering 𝜋 radians on the circle? Answer: Step 1 Compute the radius r first using the two points (7, 6) and (2, 1) 𝑟 = √(7 − 2)2 + (6 − 1)2 = √52 + 52 √2(25) = 5√2 arc 𝑠 = 𝑟𝜃 𝑠 = (5√2) × 𝜋 𝑠 = 5 × 𝜋 × √2 Question #18 The diameter of two cylinder are in the ratio of 2: 3 . What will be the ratio of their heights if their volumes are equal? Answer: Step 1 Let the ratio of height of two cylinders be 𝑘: 1 , say heights are kh and h As diameters are in the ratio of 2: 3 , let diameters be 2d and 3d Hence their volumes are 2𝑑 2 𝜋 ( ) × 𝑘ℎ = 𝜋 𝑘 ℎ𝑑 2 2 and 3𝑑 2 𝜋( ) ×ℎ = 2 9𝜋 4 ℎ𝑑 2 As two volumes are equal, we have 9𝜋 2 𝜋𝑘ℎ𝑑 2 = ℎ𝑑 4 or 9𝜋 2 ℎ𝑑 𝑘 = 4 2 = 9: 4 𝜋ℎ𝑑 Hence ratio of their heights is 9:4 Question #19 A circle has a chord that goes from 𝜋 3 to 5𝜋 12 radians on the circle. If the area of the circle is 7𝜋 , what is the length of the chord? Answer: Step 1 If you draw the radii from the centre of the circle to both ends of the chord , then you will have a triangle. Now the angle subtended at the centre by the chord will be. 5𝜋 𝜋 𝜋 − = 12 3 4 The radius of the circle may be found , since we are given it's area, by using the following. 7𝜋 𝜋𝑟 2 = 7𝜋 → 𝑟 2 = = 7 → 𝑟 = √7 𝜋 The triangle now has 2 sides (radii) = √7 𝜋 with an angle between them of 4 and a third side - the chord. To find the length of the chord , from the triangle, we need to use the cosine rule 𝑎2 = 𝑏 2 + 𝑐 2 − (2𝑏𝑐\𝑐𝑖𝑠𝜃) For this question a , is the chord , b and c , the radii and 𝜋 𝜃= 4 so 𝜋 2 2 𝑎2 = (√7) + (√7) − (2 × √7 × √7 × cos ( )) 4 𝜋 = 7 + 7 − (14 cos ( )) = 14 − (9.9899) = 4.101 4 now 𝑎2 = 4.101 ⇒ 𝑎 ( length of chord) = √4.101 ≈ 2.02 Question #20 What is the center and radius of the circle with equation (𝑥 − 5)2 + (𝑦 + 3)2 = 16 ? Answer: The circle's equation is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 Where the circle's center is (−ℎ, −𝑘) to solve the equation (𝑥 − 5)2 + (𝑥 + 3)2 = 16 As a result, the inverse of both h and k, or (5, −3), is the location of the circle's center. The radius is the sqrt16 which is 𝑟 = 4.