circle.pdf

CIRCLES An overview of conic section Definition and equation of circle Finding the equation of a circle ❖ Equation of a circle centered at (0,0) ❖ Circles with center not at the origin  Conic sections A conic section is the intersection of a plane and a cone. Observe the...

CIRCLES An overview of conic section Definition and equation of circle Finding the equation of a circle ❖ Equation of a circle centered at (0,0) ❖ Circles with center not at the origin  Conic sections A conic section is the intersection of a plane and a cone. Observe the shape of the slice that results. The angle at which the cone is sliced produces four different types of conic sections.  Circle A circle consists of all points on the plane equidistant from a fixed point called the center. The distance from the center to any point on the circle is constant and is called the radius of the circle.  The Standard Form of the Equation of a Circle Circle with center (h, k) Circle with center at the origin (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 𝑥2 + 𝑦2 = 𝑟2  Example 1: find the equation of a circle with centered at the origin and a radius of 5 units. Center (0, 0) and r = 5 𝑥 2 + 𝑦 2 = 52 𝑥 2 + 𝑦 2 = 25  Example 2: find the equation of a circle with centered at (0, 3) and with radius of 6. 2 2 2 Center (0, 3) and r = 6 (𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟 2 2 2 h=0 𝑥 + (𝑦_____) = 𝑟 k=3 𝑥 2 + (𝑦 − 3)2 = 62 2 2 𝑥 + (𝑦 − 3) = 36  Example 3: Find the equation of a circle with centered at (2, -5) and with radius of 10. 2 2 2 Center (2, -5) and r = 10 (𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟 h=2 (𝑥_____)2 +(𝑦_____)2 = 𝑟 2 k = -5 (𝑥 − 2)2 +(𝑦 + 5)2 = 102 2 2 (𝑥 − 2) +(𝑦 + 5) = 100  Example 4: Write the equation of a circle centered at (1, 5) and with radius 17 2 2 2 Center (1, 5) and r = 17 (𝑥 − ℎ) + (𝑦 − 𝑘) = 𝑟 h=1 2 2 2 (𝑥_____) +(𝑦_____) = 𝑟 k=5 (𝑥 − 1)2 +(𝑦 − 5)2 = ( 17)2 (𝑥 − 1)2 +(𝑦 − 5)2 = ( 17)2  Example 5: Find the center and the radius of the 2 2 circle 𝑥 + 𝑦 = 64 Circle with center at the origin 𝑥2 + 𝑦2 = 𝑟2 The center of the circle is (0, 0) and its radius is 𝑥 2 + 𝑦 2 = 64 8 units 64 = 8 𝑥2 + 𝑦2 = 8  Example 6: Find the center and the radius of the 2 2 circle 𝑥 + 𝑦 = 19 Circle with center at the origin 𝑥2 + 𝑦2 = 𝑟2 The center of the circle is (0, 0) and its radius is 𝑥 2 + 𝑦 2 = 19 19 units  Example 7: Find the center and the radius of 2 2 the circle 𝑥 + (𝑦 − 2) = 49 𝑥 2 + (𝑦 − 2)2 = 49 0 Center (0, 2) and r = 7 2 49 = 7 Example 8: Find the center and the radius of the circle (𝑥 + 7)2 +(𝑦 − 5)2 = 56 Center (-7, 5) and (𝑥 + 7)2 +(𝑦 − 5)2 = 56 r=2 14 -7 5 56 = 4 ∙ 14 = 2 14 CIRCLES The General Form of the Equation of a Circle Write the Equations in Standard Form Find the center and the radius of the Circle  The General Form of the Equation of a Circle 𝑥 2 + 𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 If D=-2h, E= -2k, and F= ℎ2 + 𝑘 2 − 𝑟 2 (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 𝑥 2 + 𝑦 2 + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 (𝑥 2 − 2ℎ𝑥 + ℎ2 ) + (𝑦 2 − 2𝑘𝑦 + 𝑘 2 ) = 𝑟 2 𝑥 2 − 2ℎ𝑥 + ℎ2 + 𝑦 2 − 2𝑘𝑦 + 𝑘 2 = 𝑟 2 𝑥 2 + 𝑦 2 − 2ℎ𝑥 − 2𝑘𝑦 + ℎ2 + 𝑘 2 = 𝑟 2 𝑥 2 + 𝑦 2 − 2ℎ𝑥 − 2𝑘𝑦 + ℎ2 + 𝑘 2 − 𝑟 2 = 0  Example 1: Write the general equation of a circle with center (4, -1) and a radius of 7 units. h k (4, -1) r = 7 2 2 2 (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 (𝑥 − 4) + (𝑦 + 1) = 7 2 2 (𝑥 − 4) + (𝑦 + 1) = 49 𝑥 2 − 8𝑥 + 16 + 𝑦 2 + 2𝑦 + 1 = 49 𝑥 2 − 8𝑥 + 16 + 𝑦 2 + 2𝑦 + 1 − 49 = 0 𝑥 2 + 𝑦 2 − 8𝑥 + 2𝑦 − 32 = 0  2 2 Example 2: Write 𝑥 + 𝑦 + 6𝑥 − 7 = 0 in standard form. 𝑥 2 + 6𝑥 + 𝑦 2 = 7 1 2 2 6 = 32 = 9 (𝑥 +6𝑥 + _____) + 𝑦 = 7 2 (𝑥 2 +6𝑥 + 9) + 𝑦 2 = 7 + 9 (𝑥 + 3)2 +𝑦 2 = 16  2 2 Example 3: Write 𝑥 + 𝑦 − 4𝑥 + 18𝑦 + 35 = 0 in standard form. 2 2 𝑥 − 4𝑥 + 𝑦 + 18𝑦 = −35 2 2 (𝑥 −4𝑥 + ____) + (𝑦 +18𝑦 + ____) = −35 1 1 −4 = (−2)2 = 4 18 = (9)2 = 81 2 2 (𝑥 2 −4𝑥 + 4) + (𝑦 2 +18𝑦 + 81) = −35 + 4 + 81 2 2 (𝑥 − 2) +(𝑦 + 9) = 50  Example 4: What is the center and the radius of the circle 𝑥 2 + 𝑦 2 − 6𝑥 − 10𝑦 + 18 = 0? 𝑥 2 + 𝑦 2 − 6𝑥 − 10𝑦 + 18 = 0 (𝑥 2 − 6𝑥 + ____) + (𝑦 2 −10𝑦 + ____) = −18 1 1 −6 = (−3)2 = 9 −10 = (−5)2 = 25 2 2 (𝑥 2 − 6𝑥 + 9) + (𝑦 2 −10𝑦 + 25) = −18 + 9 + 25 (𝑥 − 3)2 +(𝑦 − 5)2 = 16 16 = 4 (𝑥 − 3)2 +(𝑦 − 5)2 = 4 The center of the circle is (3, 5) and its radius is 4 units. Example 5: What is the center and the radius of the circle 4𝑥 2 + 4𝑦 2 + 12𝑥 − 4𝑦 − 90 = 0? 2 2 4𝑥 + 4𝑦 + 12𝑥 − 4𝑦 − 90 = 0 4𝑥 2 + 4𝑦 2 + 12𝑥 − 4𝑦 = 90 4 90 𝑥 2 + 𝑦 2 + 3𝑥 − 𝑦 = 4  90 (𝑥 2 + 3𝑥 + ____) + (𝑦 2 −𝑦 + ____) = 4 1 3 2 9 1 1 2 1 3 =( ) = −1 = (− ) = 2 2 4 2 2 4 2 9 2 1 90 9 1 (𝑥 + 3𝑥 + ) + (𝑦 −𝑦 + ) = + + 4 4 4 4 4 3 2 1 2 (𝑥 + ) +(𝑦 − ) = 25 2 2 25 = 5 3 2 1 2 (𝑥 + ) +(𝑦 − ) = 5 2 2 3 1 The center of the circle is (− 2 , 2) and its radius is 5 units.  CIRCLES Sketching the Graph of a Circle Problem Solving and Applications  SKETCHING THE GRAPH OF A CIRCLE Example 1: Sketch the graph of a circle with radius 4 and center at (0, 0). 5 4 3 2 2 𝑥 + 𝑦 = 16 2 1 0 -5 -4 -3 -2 -1 0 1 2 3 4 5 -1 -2 -3 -4 -5  SKETCHING THE GRAPH OF A CIRCLE Example 2: Sketch the graph of the equation (𝑥 − 2)2 +(𝑦 + 1)2 = 36. 6 5 4 Center (2, -1) and r= 6 3 2 1 0 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -1 -2 -3 -4 -5 -6 -7 -8  SKETCHING THE GRAPH OF A CIRCLE Example 3: Sketch the graph of the equation 𝑥 2 + 𝑦 2 + 10𝑥 − 4𝑦 − 8 = 12. (𝑥 2 +10𝑥 + ____) + (𝑦 2 −4𝑦 + ____) = 12 + 8 (𝑥 2 +10𝑥 + 25) + (𝑦 2 −4𝑦 + 4) = 20 + 25 + 4 10 2 2 9 (𝑥 + 5) +(𝑦 − 2) = 49 8 7 6 Center (-5, 2) r= 7 5 4 3 2 1 0 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 -1 -2 -3 -4 -5 -6  PROBLEM SOLVING AND APPLICATION Slope of the line: Given line containing two distinct points 𝑃1 𝑥1 , 𝑦1 𝑎𝑛𝑑𝑃2 𝑥2 , 𝑦2 𝑤ℎ𝑒𝑟𝑒 𝑥1 ≠ 𝑥2 , 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑚 𝑜𝑓 𝑙𝑖𝑛𝑒 𝑙 𝑖𝑠 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑎𝑠 𝑦2 − 𝑦1 𝑚1 = 𝑥2 − 𝑥1 Distance Formula: The distance between two points 𝑃1 𝑥1 , 𝑦1 and 𝑃2 𝑥2 , 𝑦2 , on an xy plane is given as 𝑑 = (𝑥2 −𝑥1 )2 + (𝑦2 −𝑦1 )2 Midpoint Formula: 𝑥2 +𝑥1 𝑦2 +𝑦1 𝑥= and 𝑦 = 2 2  PROBLEM SOLVING AND APPLICATION Example 4: Find the standard equation of the circle whose diameter has endpoints (-5, 3) and (7, 11). 𝑥2 +𝑥1 7+(−5) 2 𝑥= 2 = 2 = =1 2 𝑦2 + 𝑦1 11 + 3 14 𝑦= = = =7 2 2 2  PROBLEM SOLVING AND APPLICATION Example 4: Find the standard equation of the circle whose diameter has endpoints (-5, 3) and (7, 11). Center (1,7) and r= 2 13 𝑑= (𝑥2 −𝑥1 )2 + (𝑦2 −𝑦1 )2 (𝑥 − 1)2 + (𝑦 − 7)2 = (2 13)2 = (7 − 1)2 +(11 − 7)2 (𝑥 − 1)2 + (𝑦 − 7)2 = 52 = (6)2 +(4)2 = 36 + 16 = 52 = ±2 13 PROBLEM SOLVING AND APPLICATION Example 5: A seismological station is located at (0, -4), 4km away from a straight shoreline where the x-axis runs through. The epicenter of an earthquake was determined to be 6 km away from the station. a. Find the equation of the curve that contains the possible location of the epicenter. b. If furthermore, the epicenter was determined to be 1 km away from the shore, find its possible coordinates (round off to two decimal places).  PROBLEM SOLVING AND APPLICATION Example 5: A seismological station is located at (0, -4), 4km away from a straight shoreline where the x-axis runs through. The epicenter of an earthquake was determined to be 6 km away from the station. a. Find the equation of the curve that contains the possible location of the epicenter. Solution: Center (0, -4) and r= 6 𝑥 2 + (𝑦 + 4)2 = 36  PROBLEM SOLVING AND APPLICATION b. If furthermore, the epicenter was determined to be 1 km away from the shore, find its possible coordinates (round off to two decimal places). Solution: 𝑥 2 + (𝑦 + 4)2 = 36 Solving for x when y is 1 and -1 𝑥 = ± 11 𝑜𝑟 ± 3.31 𝑥 2 + (1 + 4)2 = 36 𝑥 2 + (−1 + 4)2 = 36 𝑥 = ± 27 𝑜𝑟 ± 5.2 𝑥 2 + (5)2 = 36 𝑥 2 + (3)2 = 36 𝑥 2 + 25 = 36 𝑥 2 + 9 = 36 (3.31, 1), (-3.31, 1), 𝑥 2 = 36 − 25 𝑥 2 = 27 (5.2, -1) and (-5.2, -1) 𝑥 2 = 11 𝑥 = ± 27 𝑥 = ± 11