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Precalculus – 2015 circle.

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Presentation on theme: "Precalculus – 2015 circle."— Presentation transcript:

1 Precalculus – 2015 circle

2 The set of all points satisfying the equation gives
x -4 4 8 r height = y-k Base = x-h Pythagorean Theorem? The set of all points satisfying the equation gives the circle with center (h,k) and radius = r .

3 Circle The set of all co-planar points equidistant from a fixed point (center). radius

4 Circle Equation: (x – h)2 + (y – k)2 = r2 r (h, k)

5 Circle r = r = 11 x – 3 = 0 x = 3 y – 5 = 0 y = 5 121 11 (3, 5) center

6 Circle r = r = 9 x + 5 = 0 x = -5 y – 2 = 0 y = 2 81 9 (-5, 2) center

7 Graph the following circle
9x2 + 36x + 9y2 - 18y - 10 = 89 9x2 + 36x + 9y2 - 18y = 9x2 + 36x + 9y2 - 18y = 99 9 (x2 + 4x + 22 ) + (y2 - 2y + (-1)2 ) = (x + 2)2 + (y - 1)2 = 16

8 Circle r = r = 4 x + 2 = 0 x = -2 y – 1 = 0 y = 1 16 4 (-2, 1) center

9 Graph the following circle
4x2 + 24x + 4y2 + 32y + 13 = 157 4x2 + 24x + 4y2 + 32y = 4x2 + 24x + 4y2 + 32y = 144 4 (x2 + 6x + 32 ) + (y2 + 8y + 42 ) = (x + 3)2 + (y + 4)2 = 61

10 Circle r = x + 3 = 0 x = -3 y + 4 = 0 y = -4 61 61 (-3, -4) center

11 Graph the following circle
5x2 - 80x + 5y2 + 20y - 34 = 106 5x2 - 80x + 5y2 + 20y = 5x2 - 80x + 5y2 + 20y = 140 5 (x2 - 16x + (-8)2 ) + (y2 + 4y + 22 ) = (x - 8)2 + (y + 2)2 = 96

12 Circle r = r = 4 6 x - 8 = 0 x = 8 y + 2 = 0 y = -2 96 4 6 (8, -2)
4 6 y + 2 = 0 (8, -2) y = -2 center (8, -2)

13 Graph the following circle
2x2 + 10x + 2y2 + 8y + 4 = 25 2x2 + 10x + 2y2 + 8y = 2x2 + 10x + 2y2 + 8y = 21 2 (x2 + 5x ) + (y2 + 4y + 22 ) = (x + )2 + (y + 2)2 =

14 Circle r = r = x + = 0 x = y + 2 = 0 y = -2 center

15 Circles Find the equation of a circle where the endpoints of a diameter are (-5,-7) and(-5,11)

16 Circles Find the equation of a circle where the endpoints of a diameter are (-4,-6) and(2,2)

17 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.

18

19 Happy Pi-Day March 14

20 Homework: Circles Worksheet


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