Questions and Answers #1 Circles

Questions and Answers Sheet 1 Circles Question #1 Along, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is 𝐽⃗. The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship b J = ( )e( r − a )/ kˆ for r  a r = 0 𝑓𝑜𝑟 ≥ 𝑎 where the radius of the cylinder is a = 5.00 cm, r is the radial distance from the cylinder axis, b is a constant equal to 600 A/m, and 𝛿 is a constant equal to 2.50 cm. (a) Let 𝐼0 be the total current passing through the entire cross section of the wire. Obtain an expression for 𝐼0 in terms of b, d, and a. Evaluate your expression to obtain a numerical value for 𝐼0 . ⃗⃗ in the region 𝑟 ≥ 𝑎 . Express (b) Using Ampere’s law, derive an expression for the magnetic field 𝐵 your answer in terms of 𝐼0 rather than b. (c) Obtain an expression for the current | contained in a circular cross section of radius r... a and centered at the cylinder axis. Express your answer in terms of 𝐼0 rather than b. ⃗⃗ in the region 𝑟 ≥ 𝑎. (d) Using Ampere’s law, derive an expression for the magnetic field 𝐵 (e) Evaluate the magnitude of the magnetic field at r- = 6, r= a, and r= 2a. Answer: (a) Consider a long, solid cylinder which has a circular cross-section of radius a and oriented with its axis in the z- direction, the current density is not constant across the cross-section of the wire and it is given by b J = ( )e( r − a )/ kˆ r where r is the distance from the center. First we need to find an expression for the total current passing through the entire cross section of the wire 𝐼0 in terms of b, 𝛿, and a. Consider a small segment, a ring with a width of dr and at distance from the center of the circle, the current in this element equals the area of the element multiplied by the current density J, that is, 𝑑𝐼 = 𝐽𝑑𝐴 the area of the element equals the area of a rectangle with a width of dr and length of 2𝜋𝑟 (the circumference of a circle with a radius of r), that is, 𝑑𝐴 = 2𝜋𝑟𝑑𝑟 b r thus, dl = ( )e ( r − a )/  2 rdr = 2 be ( r − a )/ dr the total current passing throught the wire equals the integration over the whole wire, from r = 0 to r = a, that is, 𝑎 𝑎 𝐼0 = ∫0 𝑑𝐼 = 2𝜋𝑏 ∫0 𝑒 (𝑟−𝑎)/𝛿 𝑑𝑟 0 = 2𝜋𝑏𝛿𝑒 (𝑟−𝑎)/𝛿|𝑎 = 2𝜋𝑏𝛿(1 − 𝑒 −𝑎/𝛿 ) I0 = 2 b (1 − e)−a / the radius of the cylinder is a=5.00,b is a constant equal to 600 A/m, and d is a constant equal to 2,50 cm, substitute with the givens to get, 𝐼 0 = 2𝜋(600𝐴/𝑚)(0.025𝑚)(1 − 𝑒 (0.050/0.025) ) = 81.5𝐴 𝐼 0 = 81.5𝐴 Question #2 How to determine the arc length of ellipse? I want to determine the length of an arc from the ellipse in the picture below: 𝜃 = 50∘ a=3.05cm b=2.23cm d=?cm How can I determine the length of d? Answer: Let a=3.05, b=2.23 . Then a parametric equation for the ellipse is x=a 𝑐𝑜𝑠 t, y=b 𝑠𝑖𝑛 . When t=0 the point is at (a,0)=(3.05,0) , the starting point of the arc on the ellipse whose length you seek. Now it's important to realize that the parameter t is not the central angle, so you need to get the value of t 𝑦 which corresponds to the top end of your arc. At that end you have = tan 5 0 (degrees). And in terms of 𝑥 t you have 𝑦 𝑥 𝑏 𝑎 = ( ) 𝑡𝑎𝑛. Solving for t then gives 𝑡 = 𝑡1 = 𝑎𝑟𝑐 tan ( tan 5 0). 𝑎 [note I'd suggest using radians here, replacing the 50 by 𝑏 5𝜋 18 .] For the arclength use the general formula of integrating √𝑥 ′2 + 𝑦 ′2 for t in the desired range. In your case 𝑥 ′ = −𝑎 sin 𝑡 , 𝑦 ′ = 𝑏 𝑐𝑜𝑠 so that you are integrating √𝑎2 sin2 𝑡 + 𝑏 2 cos 2 𝑡 with respect to t from 0 to the above 𝑡1 . There not being a simple closed form for the antiderivative (it's an ''elliptic integral), the simplest approach now would be to do the integral numerically. This seems the more appropriate in your problem as you only know a,b to two decimals, apparently. * When I did this numerically on maple I got about 2.531419 for the arclength. Question #3 Find the points on the ellipse 4𝑥 2 + 𝑦 2 = 4 that are farthest away from the point (1,0). Answer: Question #4 Find an equation for the surface obtained by rotating the parabola 𝑦 = 𝑥 2 about the y-axis. Answer: Question #5 Two circles have a scale factor of 3:2. We need to find the ratio of areas of these circles. 1) 9:4 2) 13:12 3) 12:26 4) 3:2 Answer: Let, 𝑟1 and 𝑟2 be the radii of two circles. According to the question, 𝑟1 3 = 𝑟2 2 Ratio of areas of both circles 𝜋𝑟12 𝜋𝑟22 𝑟 2 3 2 9 2 4 = ( 1) = ( ) = 𝑟2 First option is the correct answer. Question #6 To find the equation of two circles with centers A (1, 1) and B (9, 7) which have an equal radius and touch each other externally and also the equation of common tangents to these circles. Answer: Step 1 Consider the diagram of the given situation: Here AP is the radius of the circle with center A BP is the radius of the circle with center B Given AP=BP=r 𝐴𝐵 = √(9 − 1)2 + (7 − 1)2 𝐴𝐵 = √64 + 36 𝐴𝐵 = √100 AP+BP=10 r+r=10 2r=10 r=5 Therefore the radius of the circles is 5 Equation of circle with center C(h, k) and radius r is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 Here Equation of circle with center A(1, 1) and radius 5 is (𝑥 − 1)2 + (𝑦 − 1)2 = 25 Equation of circle with center A(9, 7) and radius 5 is (𝑥 − 9)2 + (𝑦 − 7)2 = 25 Step 2 Here point P divides the line segment into 1:1 ratio, So the coordinate of point P is 𝑃=( 1+9 7+1 2 , 2 10 8 ) = ( , ) = (5,4) 2 2 Line AP is perpendicular to tangent line, therefore 𝑚1 × 𝑚2 = −1 4−1 × 𝑚2 = −1 5−1 4 𝑚2 = − 3 4 Equation of tangent line is 𝑦 = − 𝑥 + 𝑐 and it passes through point P(5, 4) so 3 4 𝑦=− 𝑥+𝑐 3 4 4=− ×5+𝑐 3 𝑐 = 4+ 𝑐= 32 20 3 3 4 32 3 3 Hence the required equation of common tangent is 𝑦 = − 𝑥 + Question #7 Write the equation of a line passing through the centers of the two circles (𝑥 − 3)2 + (𝑦 + 2)2 = 16 and (𝑥 − 6)2 + (𝑦 + 1)2 = 25 Answer: Step 1 Comparing with standard equation of a circle ic (𝑥−. ℎ)2 + (𝑦 − 𝑘)2 = 𝑠 2 where (h,k): center of circle From equation (1), centre: (ℎ1 , 𝑘1 ) = (3, −2) Froma equation (2), centre: (ℎ2 , 𝑘2 ) = (6, −1) Step 2 Since equation of a line bassing through two paints (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is given by, 𝑦 − 𝑦1 = ( 𝑦2 −𝑦1 𝑥2 −𝑥1 ) (𝑥 − 𝑥1 ) ∴ equation of siqusid line, 𝑦 − 𝑘1 = ( 𝑦+2 =( 𝑘2 −𝑘1 ℎ2 −ℎ1 −1+2 3 ) (𝑥 − ℎ1 ) ) (𝑥 − 3) 1 𝑦 + 2 = (𝑥 − 3) 𝑦= 𝑦= 3 1𝑥−1−2 3 1𝑥−3 3 '' Question #8 James has two circles, Circle A and Circle B. Use the information to answer the questions that follow. Circle A: The equation for circle A is 𝑥 2 + 𝑦 2 + 6𝑥 − 4𝑦 − 12 = 0 Circle B: center (0, -4) with a radius of 3 A) Complete the square in order to find the center and radius of Circle A. Show your work and state the center and radius of the circle. B) Does the point (√8, −3) lie on Circle B? Support your answer with evidence graphically or algebraically. C) Are Circle A and Circle B similar? Yes or no and prove your answer. Answer: a) Circle A 𝑥 2 + 𝑦 2 + 6𝑥 − 4𝑦 − 12 = 0 (𝑥 2 + 6𝑥) + (𝑦 2 − 4𝑦) − 12 = 0 (𝑥 2 + 6𝑥 + 9 − 9) + (𝑦 2 − 4𝑦 + 4 − 4) − 12 = 0 (𝑥 + 3)2 + (𝑦 − 2)2 − 25 = 0 (𝑥 + 3)2 + (𝑦 − 2)2 = 25 Centre 𝑥 + 3 = 0 ⇒ 𝑥 = −3 𝑦−2=0⇒𝑦=2 𝑟 2 = 25 ⇒ 𝑟 = 5 Centre (-3, 2) radius =5 B) Circle B Equation (𝑥 − 0)[2} + (𝑦 + 4)2 = 32 = 9 Check (√8, −3) 2 (√8) + (−3 + 4)2 = 9 8+1=9 ∵ Point (√8, −3) Satisfy the equation of circle B hence point (√8, −3) lie on Circle B. C) Circle A and Circle B ore similar (because all circle are of the same shape hence all circle are similar). Question #9 The radius of ach circle is 4cm. What is the total area of the 6 identical circles? (Take: =3.14 ) Answer: The radius of each circle is 4 cm. So, area of each circle is: 𝐴 = 𝜋(4)2 =50.24 Therefore, the area of 6 identical circles is 6𝐴 = 6(50.24) = 301.44𝑐𝑚2 Question #10 The diagram shows a running track. It is made up of a rectangle and two semi circles. A student ran around the track 7 times. What is the total distance in m, he ran? Use 𝜋 = 22/7 Answer: Step 1 Given: diameter(D) of semicircle =7m, length(l) of rectangle =15m in one round total distance(d) covered is given by 𝑑 = 𝜋(𝐷) + 2(1) Step 2 so, distance covered in seven rounds will be 22 𝑑 = 7(𝜋𝐷 + 2(1)) (∵ 𝜋 = , 𝐷 = 7,1 = 15) 7 = 7𝜋𝐷 + 141 22 = 7 ( ) (7) + 14(15) 7 =154+210 =364m hence, total distance covered is equal to 364m Question #11 Suppose you have three concentric circles and are looking at the arc lengths of a 80∘ angle of each. If the radii of the circles have a ratio of 2:3:5 what is the ratio of the arc lengths? Answer: Step 1 Given: Arc lenghs of each circle =80 Ratio of radii of the circles =2:3:5 Step 2 All circles are similar. In similar figures the ratio of corresponding sides are proportional. So, ratio of are lengths is same as the ratio of radii. so, ratio of are lengths is given by: ⇒ 2: 3: 5 Question #12 Calculate the volume of this shape. Leave your answer in terms of 𝜋. Answer: Step 1 This is the shape of a cylinder: Now, formula for calculating the volume of the cylinder is: 𝑉 = 𝜋𝑟 2 ℎ Given figure has two circles one above and one at the bottom; These circles have equal areas; We are given the diameter of the circles as d=14cm, we can find the radius (r) as r=7cm Step 2 From the labelled figure, we get Radius (r)=7cm Height (h)=7cm Putting the values in the formula ,we get: 𝑉 = 𝜋(7)2 (7)𝑐𝑚3 𝑉 = 343𝜋𝑐𝑚3 Leaving it in terms of 𝜋: Volume of Cylinder = 343𝜋𝑐𝑚3 Question #13 What is the radius of the circle for the equation below? Round to the tents place. 𝑥 2 + 𝑦 2 + 8𝑥 + 10𝑦 + 12 = 0 a) 39.0 b) 5.4 c) 29.0 d) 6.2 Answer: Step 1 This question is about application of circles Step 2 To find radius we need equation in (𝑥 − 𝑎)2 + (𝑦 − 6)2 = 𝑟 2 ∴ here 𝑥 2 + 𝑦 2 + 8𝑥 + 10𝑦 + 12 = 0 𝑥 2 + 8𝑥 + 16 + 𝑦 2 + 10𝑦 + 25 = 16 + 25 − 12 𝑥 2 + 2(4𝑥) + 16 + 𝑦 2 + 2(54) + 25 = 43 − 12 (𝑥 + 4)2 + (𝑦 + 5)2 = 29 ∴ 𝑟 2 = 29 𝑟 = √29 𝑟 =5⋅4 Question #14 To find: To explain why quadrilateral ABCD is a square by using the following condition, ''Two congruent intersecting circles B and D (not shown) have a line (segment) of centers 𝐵𝐷 and a common chord 𝐴𝐶 that are congruent.'' Answer: Given that circles B and D are congruent. Also centers 𝐵𝐷 and a common chord 𝐴𝐶 that are congruent. That is ≅ 𝐴𝐶 The diagrammatic representation is given below, Form the above circles AB=BC . Since radius of the circle B is same for all points on the circle. Similarly, AD=DC Also the circles B and D are congruent. Therefore, AB=BC=CD=DA Then all the four sides are equal then ABCD is said to be rhombus. Also given that the diagonals of ABCD are congruent. That is 𝐵𝐷 ≅ 𝐴𝐶 Hence ABCD is a square. So, ABCD is a square. Final statement: The ABCD is a square. Question #15 Circle A and circle B are congruent. CD is the chord of both circles. The radius is 13 in and CD=24 in. What is the length of AB? a) 13 in b) 20 in c) 18 in d) 10 in Answer: Step 1 Let Line CD meet AB at point E. We know that perpendicular from center to chord bisect the chord, therefore, CE=DE=12 inches Now, in triangle ACE, BCE AC=CB {radius of both circles are equal}, side CE is common to both and angle AEC=angle CEB { both are of 90 degrees} Therefore, triangle ACE is congruent to triangle BCE Therefore, AE=EB Step 2 Now, by Pythagorus theorem 𝐴𝐸 = √(𝐴𝐶)2 − (𝐶𝐸)2 𝐴𝐸 = √(13)2 − (12)2 𝐴𝐸 = √169 − 144 = 5 𝐴𝐵 = 2𝐴𝐸 = 10 inches Step 3 Ans: 10 inches Question #16 Find the coordinates of the points of intersection of each of the following pairs of circles. a. 𝑥 2 + 𝑦 2 − 3𝑥 + 5𝑦 − 4 = 0 and 𝑥 2 + 𝑦 2 − 𝑥 + 4𝑦 − 7 = 0 Answer: Step 1 Since you have asked multiple questions in single request so we will be answering only first question. Given: a) a. 𝑥 2 + 𝑦 2 − 3𝑥 + 5𝑦 − 4 = 0 ⇒ 1 𝑥 2 + 𝑦 2 − 𝑥 + 4𝑦 − 7 = 0 ⇒ 1 Step 2 −2𝑥 + 𝑦 + 3 = 0 ⇒ 𝑦 = −3 + 2𝑥 ⇒ 3 Substitute value of y in 1, 𝑥 2 + (−3 + 2𝑥)2 − 3𝑥 + 5(−3 + 2𝑥) − 4 = 0 Step 3 𝑥 2 + (9 − 12𝑥 + 4𝑥 2 ) − 3𝑥 + 5(2𝑥 − 3) − 4 = 0 𝑥 2 + 9 − 12𝑥 + 4𝑥 2 − 3𝑥 + 10𝑥 − 15 − 4 = 0 ⇒ 5𝑥 2 − 5𝑥 − 10 = 0 ⇒ 𝑥2 − 𝑥 − 2 = 0 ⇒ 𝑥 2 − 2𝑥 + 𝑥 − 2 = 0 ⇒ 𝑥(𝑥 − 2) + 1(𝑥 − 2) = 0 ⇒ (𝑥 − 2)(𝑥 + 1) = 0 ⇒ 𝑥 = −1, 𝑥 = 2 Step 4 but these values of x in 3, 𝑦 = −3 + 2(−1) = −5 𝑦 = −3 + 2(2) = 1 ⇒ points of intersection are (-1,-5) and (2,1) Question #17 Explain it clearly and easy to understand. Show step-by-step. (In this in case, 𝜋 = 3.1416 ) Round the answer to the nearest tenth. 1 There are two circles. The diameter of the first is 12 in. and the area of the second is 2 times the area of 4 the first. What is the diameter of the second circle? Answer: Step 1 Let d and 𝐴1 be the diameter and area of first circle and D and 𝐴2 be diameter and area of second circle. 1 Then, according to given condition, d=12 and 𝐴2 = 2 𝐴1 4 9 𝐴2 = 𝐴1 4 We know that area of circle is 𝜋(𝑟𝑎𝑑𝑖𝑢𝑠)2 , that is 𝜋(𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2)2 Step 2 9 Therefore, 𝐴2 = 𝐴1 𝐷 2 9 𝑑 2 2 4 2 4 𝜋( ) = 𝜋( ) 9 𝐷2 = 𝑑 2 3 4 𝐷= 𝑑 2 3 𝐷 = (12) = 24 in. 2 Step 3 Answer: Diameter of second circle is 24 in. Question #18 Identify the center and radius of the following circles, as well as graph each circle: (𝑥 − 3)2 + (𝑦 + 8)2 = 144 𝑥 2 + (𝑦 + 2)2 = 56.25 Answer: Step 1 Given: equation of circles 1) (𝑥 − 3)2 + (9 + 8)2 = 144 2) 𝑥 2 + (9 + 2)2 = 56.15 Equation of circle centred at (h,k) radius is (𝑥 − 𝑛)2 + (𝑦 − 𝑘)2 = 𝑟 2 Step 2 Solution 1) (𝑥 − 3)2 + (𝑦 + 8)2 = 144 here 122 = 144 (𝑥 − 3)2 + (𝑦 + 8)2 = 122 2 (𝑥 − 3)2 + (𝑦 − (−8)) = 122 comparing with the equation (𝑥 − 𝑛)2 + (𝑦 − 𝑘)2 = 𝑟 2 (ℎ, 𝑘) = (3, −8)𝑟 = 12 centre =(3,-8) Radius =12 2) 𝑥 2 + (𝑦 + 2)2 = 56.25 2 (𝑥 − 0)2 + (𝑦 − (−2)) = (7.5)2 (7.5)2 = 56.25 comparing with the equation, (𝑥 − 𝑛)2 + (𝑦 − 𝑘)2 = 𝑟 2 (𝑛, 𝑘) = (0, −2)𝑟 = 7.5 centre =(0,-2) Radius =7.5 Question #19 Identify the center and radius of the circles 1) (𝑥 − 1)2 + (𝑦 + 3)2 = 4 2) (𝑥 − 2)2 + (𝑦 − 1)2 = 16 3) 𝑥 2 + (𝑦 − 3)2 = 14 4) 𝑥 2 + 4𝑥 + 𝑦 2 = 32 Answer: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 (h, k) is the center and r is the radius of the circle. 1. (𝑥 − 1)2 + (𝑦 + 3)2 = 4 2 (𝑥 − 1)2 + (𝑦 − (−3)) = 22 (1, -3) is the center and 2 is the radius of the circle. 2. (𝑥 − 2)2 + (𝑦 − 1)2 = 16 (𝑥 − 2)2 + (𝑦 − 1)2 = 42 (2, 1) is the center and 4 is the radius of the circle. 3. 𝑥 2 + (𝑦 − 3)2 = 14 2 (𝑥 − 0)2 + (𝑦 − 3)2 = (√14) (0, 3) is the center and √14 is the radius of the circle. 4. 𝑥 2 + 4𝑥 + 𝑦 2 = 32 𝑥 2 + 4𝑥 + 𝑦 2 + 4 = 32 + 4 (𝑥 2 + 4𝑥 + 4) + 𝑦 2 = 36 (𝑥 + 2)2 + (𝑦 − 0)2 = 62 2 (𝑥 − (−2)) + (𝑦 − 0)2 = 62 (-2, 0) is the center and 6 is the radius of the circle. Question #20 The areas of two circles are 36𝜋 and 64𝜋 diameters? What is the ratio of the of the circumferences? Answer: Step 1 Area of circle is given by 𝜋𝑟 2 𝐴1 = 36 𝜋 𝜋𝑟12 = 36 𝜋 𝑟1 = 6 Step 2 𝐴2 = 64 𝜋 𝜋𝑟22 = 64 𝜋 𝑟2 = 8 Circumference, 𝐶1 = 2𝜋𝑟1 𝐶1 = 2𝜋 × 6 𝐶1 = 12 𝜋 𝐶2 = 2𝜋𝑟2 = 16 𝜋 𝐶1 : 𝐶2 = 12𝜋: 16 𝜋 =3∶ 4