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Jeff Bivin -- LZHS
Precalculus – 2015
Jeff Bivin -- LZHS
y
x
-4 4
4
8
2 2 2( ) ( )x h y k r
The set of all points satisfying the equation gives the circle with center (h,k) and radius = r .
Base = x-h
height = y-k
h
kPythagorean Theorem?
r
Jeff Bivin -- LZHS
Circle
The set of all co-planar points equidistant from a fixed point (center).
Jeff Bivin -- LZHS
CircleEquation: (x – h)2 + (y – k)2 = r2
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
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
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
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
Circles• Find the equation of a circle where the
endpoints of a diameter are
(-5,-7) and(-5,11)
Jeff Bivin -- LZHS
Circles
• Find the equation of a circle where the endpoints of a diameter are
(-4,-6) and(2,2)
Jeff Bivin -- LZHS
Circles
• Find the standard form of the equation of a circle that contains the points (9,4), (1,14) and (9,14).
• Use the idea that the perpendicular bisector of any chord of a circle passes through the center of the circle. Since the given points are on the circle, any pair of the three points are the endpoints of a line segment that is a chord of the circle.
• Find the perpendicular bisector of 2 of the chords. Their intersection will be the center of the circle. Distance from center to any of the points = radius.
2 25 9 41x y
Jeff Bivin -- LZHS
Homework:
Circles Worksheet
