University:
North Central College
Course:
MATH 140 | Precalculus
Academic year:

2019
Views:

100

Pages:

4
Author:

Sheha A.

9.1-9.3 Quiz Review Name: Hour: Find the standard form of the equation of the circle with the given characteristics. 1. Center: (0, 0); point on circle: (8, -15) x^2 + y^2 = r^2 (-15)^2 + 8^2 = r^2 225 + 64 = r^2 r^2 = 289 x^2 + y^2 = 289 2. Endpoints of diameter: (-2, 3) and (6, -5) Center: ((-2+6)/2, (3-5)/2) = (2, -1) (x - h)^2 + (y - k)^2 = r^2 (6-2)^2 + (-5+1)^2 = r^2 16 + 16 = r^2 32 = r^2 (x - 2)^2 + (y + 1)^2 = 32 Write the equation of the circle in standard form. Sketch the circle and then identify its center, radius, x-intercepts, and y-intercepts. 3. x^2 + y^2 + 4x + 6y - 3 = 0 x^2 + 4x + 4 + y^2 + 6y + 9 = 3 + 4 + 9 (x + 2)^2 + (y + 3)^2 = 16 Center: (-2, -3) Radius: 4 x-intercepts: (x + 2)^2 + 3^2 = 16 (x + 2)^2 = 7 x + 2 = ±√7 x = -2 ± √7 y-intercepts: 2^2 + (y + 3)^2 = 16 (y + 3)^2 = 12 y + 3 = ±2√3 y = -3 ± 2√3 4. 4x^2 + 4y^2 + 32x - 24y + 51 = 0 4(x^2 + 8x) + 4(y^2 - 6y) = -51 4(x^2 + 8x + 16) + 4(y^2 - 6y + 9) = -51 + 64 + 36 4(x + 4)^2 + 4(y - 3)^2 = 49 (x + 4)^2 + (y - 3)^2 = 49/4 Center: (-4, 3) Radius: 7/2 x-intercepts: (x + 4)^2 + 9 = 49/4 (x + 4)^2 = 13/4 x + 4 = ±√13/2 x = -4 ± √13/2 y-intercepts: None