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MATHPOWER TM 12, WESTERN EDITION 3.3.1 3.3 Chapter 3 Conics.

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Presentation on theme: "MATHPOWER TM 12, WESTERN EDITION 3.3.1 3.3 Chapter 3 Conics."— Presentation transcript:

1 MATHPOWER TM 12, WESTERN EDITION Chapter 3 Conics

2 Developing the Standard Forms of the Equation of a Circle P(x, y) O (0, 0) Note: OP is the radius of the circle. The standard form of the equation of a circle with its centre at the origin (0, 0) is x 2 + y 2 = r

3 C(h, k) P(x, y) This is the standard form of the equation of a circle with the centre at (h, k) Developing the Standard Forms of the Equation of a Circle

4 Determine the equation of a circle with centre C(-5, 2) and passing through the point P(-8, 7). From the general form: (x - h) 2 + (y - k) 2 = r 2 (x - (-5)) 2 + (y - 2) 2 = r 2 (x + 5) 2 + (y - 2) 2 = r 2 Substitute the values of h and k from C(-5, 2): Use the point P(-8, 7) to find the value of r 2 : (x + 5) 2 + (y - 2) 2 = r 2 (-8 + 5) 2 + (7 - 2) 2 = r = r 2 34 = r 2 Therefore, the equation of the circle in standard form is (x + 5) 2 + ( y - 2) 2 = 34. (h, k) (x, y) Finding the Equation of a Circle 3.3.4

5 (x + 5) 2 + (y - 2) 2 = 34 x x y 2 - 4y + 4 = 34 x 2 + y x - 4y + 29 = 34 x 2 + y x - 4y - 5 = 0 Write the following equation in general form: The general form of the equation is Ax 2 + Cy 2 + Dx + Ey + F = 0. (x + 5) 2 + (y - 2) 2 = Writing the General Form of the Equation of a Circle

6 Find the centre and the radius of each circle: 1. x 2 + y 2 - 8x + 10y - 14 = 0 x 2 + y 2 - 8x + 10y - 14 = 0 (x 2 - 8x + _____ ) + (y y + _____) = 14 + _____ + _____ To find the centre and radius, write the equation in standard form. To do this, you must complete the square: The centre is (4, -5) and the radius is x 2 + 3y 2 + 6x + 12y + 5 = 0 (3x 2 + 6x) + (3y y) = -5 3(x 2 + 2x + _____) + 3(y 2 + 4y + _____) = -5 + _____ + _____ The centre is (-1, -2) and the radius is (x - 4) 2 + (y + 5) 2 = (x + 1) 2 + 3(y + 2) 2 = Finding the Centre and the Radius

7 Using a Graphing Calculator Graph: (x - 3) 2 + (y - 4) 2 = 16 Your calculator will only graph a function, therefore, you must write the equation in the form y =. Make sure that you use a ZSquare graphing window. You can also use the Draw circle command on your TI-83: Press [2nd][PRGM] 9 and enter the following: Circle (3, 4, 4) 3.3.7

8 Using your graphing calculator, graph the following equations: a) x 2 + y 2 = 16 b) 4x 2 + y 2 = 16 c) 0.5x 2 + y 2 = 16 d) Ax 2 + y 2 = 16, when A = 0 x 2 + y 2 = 16 4x 2 + y 2 = x 2 + y 2 = 16 4x 2 + y 2 = 16 x 2 + y 2 = x 2 + y 2 = 16 4x 2 + y 2 = 16 x 2 + y 2 = 16 Ax 2 + y 2 = 16, when A = Using a Graphing Calculator

9 d) x 2 + 4y 2 = 16 e) x y 2 = 16 f) x 2 + Cy 2 = 16, when C = 0 x 2 + y 2 = 16 x 2 + 4y 2 = 16 x 2 + y 2 = 16 x y 2 = 16 x 2 + 4y 2 = 16 x 2 + y 2 = 16 x y 2 = 16 x 2 + Cy 2 = 16, when C = 0 Using a Graphing Calculator [contd] Using your graphing calculator, graph the following equations:

10 The values of A and C affect the graph of the circle by either a vertical or horizontal compression or expansion: A > 1 results in a horizontal compression. 0 < A < 1 results in a horizontal expansion. A = 0 results in a pair of horizontal parallel lines. C > 1 results in a vertical compression. 0 < C < 1 results in a vertical expansion. C = 0 results in a pair of vertical parallel lines. Conclusions: Using a Graphing Calculator [contd]

11 Pages 141 and 142 A 1-25 odd, B 36-45, 49, 50, 51, 54, 56, 58 (graph), Suggested Questions:


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