University:
Cornell University
Course:
MATH 1101 | Calculus Preparation
Academic year:

2020
Views:

405

Pages:

3
Author:

Nabila N.

9.1 Day 1 WS - Circles Name: Hour: Find the standard form of the equation of the circle with the given characteristics. 1. Center: (3, 7); point on circle: (1, 0) (x-3)² + (y-7)² = r² (1-3)² + (0-7)² = r² (-2)² + (-7)² = 53 = r² (x-3)² + (y-7)² = 53 2. Center (-3, -1); diameter: 2√7 r = √7 (x-h)² + (y-k)² = r² (x+3)² + (y+1)² = 7 Write the equation of the circle in standard form. Sketch the circle and then identify its center, radius, x-intercepts, and y-intercepts. 3. x² + y² - 10x - 6y + 25 = 0 x² - 10x + 25 + y² - 6y + 9 = -25 + 25 + 9 (x-5)² + (y-3)² = 9 [Image of a circle with center (5,3) and radius 3] Center: (5,3) Radius: 3 x-intercepts: (5,0) y-intercepts: None (x-5)² + (0-3)² = 9 (x-5)² = 0 x-5 = 0 x = 5 4. 4x² + 4y² + 12x - 24y + 41 = 0 4(x² + 3x) + 4(y² - 6y) = -41 x² + 3x + 9/4 + y² - 6y + 9 = -41/4 + 9/4 + 9 (x+3/2)² + (y-3)² = 1 [Image of a circle with center (-3/2, 3) and radius 1] Center: (-3/2, 3) Radius: 1 x-intercepts: None y-intercepts: None (x+3/2)² + (0-3)² = 1 (x+3/2)² = -8 (0+3/2)² + (y-3)² = 1 (y-3)² = -5/4 8. 9x² + 9y² + 54x - 36y + 17 = 0 9(x² + 6x) + 9(y² - 4y) = -17 x² + 6x + 9 + y² - 4y + 4 = -17/9 + 9 + 4 (x+3)² + (y-2)² = 100/9 [Image of a circle with center (-3,2) and radius 10/3] Center: (-3,2) Radius: 10/3 x-intercepts: (x+3)² + (0-2)² = 100/9 (x+3)² = 64/9 x + 3 = ±8/3 x = -3 ± 8/3 = -1/3, -17/3 9. An earthquake was felt up to 81 miles from its epicenter. You were located 60 miles west and 45 miles south of the epicenter. a. Let the epicenter be at the point (0,0). Find the standard equation that describes the outer boundary of the earthquake. x² + y² = 81² b. Would you have felt the earthquake? Yes c. How far were you from the outer boundary of the earthquake? Distance from epicenter to You = √(60² + 45²) = 75 81 - 75 = 6 miles Find the equation of the circle in standard form that satisfies the given conditions: The circle has a diameter with endpoints (2, 3) and (-4, 11) Center: ((2+(-4))/2, (3+11)/2) = (-1, 7) Radius = √((2-(-1))² + (3-7)²) = √(3² + (-4)²) = √25 = 5 (x+1)² + (y-7)² = 25 circle with center (1, 7) and radius 5 Find the equation of the circle in standard form that satisfies the given conditions: The circle has its center at (-2, 3) and is tangent to the y-axis r = 2 h = -2 k = 3 (x-(-2))² + (y-3)² = 2² (x+2)² + (y-3)² = 4 circle with center (2, 3) and radius 2 Find the equation of the circle in standard form that satisfies the given conditions: The circle has its center at (7, 11) and is tangent to the x-axis r = 11, h = 7, k = 11 (x-7)² + (y-11)² = 11² (x-7)² + (y-11)² = 121 circle with center (7, 11) and radius 11