10-2: conic sections - circles
DESCRIPTION
10-2: Conic Sections - Circles. circle. Circle. The set of all co-planar points equidistant from a fixed point (center). radius. Circle. Equation: (x – h ) 2 + (y – k ) 2 = r 2. (x – h ) 2 + (y – k ) 2 = r. (x, y). r. (h, k). Circle. (x – 3) 2 + (y – 5) 2 = 121. - PowerPoint PPT PresentationTRANSCRIPT
Jeff Bivin -- LZHS
10-2: Conic Sections - Circles
Jeff Bivin -- LZHS
Circle
The set of all co-planar points equidistant from a fixed point (center).
Jeff Bivin -- LZHS
Circle
Equation: (x – h)2 + (y – k)2 = r2
(x – h)2 + (y – k)2 = r
Jeff Bivin -- LZHS
Circle
x – 3 = 0
y – 5 = 0
x = 3
y = 5
r =r = 11
(x – 3)2 + (y – 5)2 = 121
Jeff Bivin -- LZHS
Circle (x + 5)2 + (y – 2)2 = 81
x + 5 = 0
y – 2 = 0
x = -5
y = 2
r =r = 9
Jeff Bivin -- LZHS
Graph the following circle9x2 + 36x + 9y2 - 18y - 10 = 89
9x2 + 36x + 9y2 - 18y = 89 + 10
(x + 2)2 + (y - 1)2 = 16
9x2 + 36x + 9y2 - 18y = 99
9
(x2 + 4x + 22 ) + (y2 - 2y + (-1)2 ) = 11 + 4 + 1
Remember when completing the
square, the coefficient of the
squared term must be 1
Jeff Bivin -- LZHS
Circle (x + 2)2 + (y – 1)2 = 16
x + 2 = 0
y – 1 = 0
x = -2
y = 1
r =r = 4
Jeff Bivin -- LZHS
Graph the following circle4x2 + 24x + 4y2 + 32y + 13 = 157
4x2 + 24x + 4y2 + 32y = 157 - 13
(x + 3)2 + (y + 4)2 = 61
4x2 + 24x + 4y2 + 32y = 144
4
(x2 + 6x + 32 ) + (y2 + 8y + 42 ) = 36 + 9 + 16
Jeff Bivin -- LZHS
Circle (x + 3)2 + (y + 4)2 = 61
x + 3 = 0
y + 4 = 0
x = -3
y = -4
r =
Jeff Bivin -- LZHS
Graph the following circle5x2 - 80x + 5y2 + 20y - 34 = 106
5x2 - 80x + 5y2 + 20y = 106 + 34
(x - 8)2 + (y + 2)2 = 96
5x2 - 80x + 5y2 + 20y = 140
5
(x2 - 16x + (-8)2 ) + (y2 + 4y + 22 ) = 28 + 64 + 4
Jeff Bivin -- LZHS
Circle (x - 8)2 + (y + 2)2 = 96
x - 8 = 0
y + 2 = 0
x = 8
y = -2
r =r = 4 6
Jeff Bivin -- LZHS
2
21
4
252
2
5
(x2 + 5x + ) + (y2 + 4y + 22 ) = + + 4
Graph the following circle2x2 + 10x + 2y2 + 8y + 4 = 25
2x2 + 10x + 2y2 + 8y = 25 - 4
(x + )2 + (y + 2)2 =
2x2 + 10x + 2y2 + 8y = 21
2
2
5
4
83
Jeff Bivin -- LZHS
Circle (x + )2 + (y + 2)2 =
x + = 0
y + 2 = 0
x =
y = -2
r =
r =
2
5
4
83
2
5
2
5
2,2
5
2,2
5
4
83
2
832
83
Jeff Bivin -- LZHS
Find the equation of the circle such that the endpoints of a diameter are (2,7) and (-6, 15).
Center: use midpoint formula
radius: use distance formula
(h,k) = 6 2 15 7, ( 2,11)
2 2
radius 2 2( 2 2) (11 7)
16 16
322 32r
2 2( 2) ( 11) 32x y