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Chapter 3 Conics 3.3 The Circle MATHPOWERTM 12, WESTERN EDITION 3.3.1

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**Developing the Standard Forms of the Equation of a Circle**

Note: OP is the radius of the circle. P(x, y) O(0, 0) The standard form of the equation of a circle with its centre at the origin (0, 0) is x2 + y2 = r2. 3.3.2

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**Developing the Standard Forms of the Equation of a Circle**

P(x, y) C(h, k) This is the standard form of the equation of a circle with the centre at (h, k). 3.3.3

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**Finding the Equation of a Circle**

Determine the equation of a circle with centre C(-5, 2) and passing through the point P(-8, 7). (h, k) (x, y) From the general form: (x - h)2 + (y - k)2 = r2 Substitute the values of h and k from C(-5, 2): (x - (-5))2 + (y - 2)2 = r2 (x + 5)2 + (y - 2)2 = r2 (x + 5)2 + (y - 2)2 = r2 (-8 + 5)2 + (7 - 2)2 = r2 = r2 34 = r2 Use the point P(-8, 7) to find the value of r2: Therefore, the equation of the circle in standard form is (x + 5)2 + ( y - 2)2 = 34. 3.3.4

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**Writing the General Form of the Equation of a Circle**

The general form of the equation is Ax2 + Cy2 + Dx + Ey + F = 0. Write the following equation in general form: (x + 5)2 + (y - 2)2 = 34 (x + 5)2 + (y - 2)2 = 34 x2 + 10x y2 - 4y + 4 = 34 x2 + y2 + 10x - 4y + 29 = 34 x2 + y2 + 10x - 4y - 5 = 0 3.3.5

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**Finding the Centre and the Radius**

Find the centre and the radius of each circle: x2 + y2 - 8x + 10y - 14 = 0 To find the centre and radius, write the equation in standard form. To do this, you must complete the square: x2 + y2 - 8x + 10y - 14 = 0 (x2 - 8x + _____ ) + (y2 + 10y + _____) = 14 + _____ + _____ 16 25 16 25 (x - 4)2 + (y + 5)2 = 55 The centre is (4, -5) and the radius is 7.4. x2 + 3y2 + 6x + 12y + 5 = 0 (3x2 + 6x) + (3y y) = -5 3(x2 + 2x + _____) + 3(y2 + 4y + _____) = -5 + _____ + _____ 1 4 3 12 3(x + 1)2 + 3(y + 2)2 = 10 The centre is (-1, -2) and the radius is 3.3.6

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**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

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**Using a Graphing Calculator**

Using your graphing calculator, graph the following equations: a) x2 + y2 = 16 b) 4x2 + y2 = 16 c) 0.5x2 + y2 = 16 d) Ax2 + y2 = 16, when A = 0 x2 + y2 = 16 Ax2 + y2 = 16, when A = 0 x2 + y2 = 16 x2 + y2 = 16 x2 + y2 = 16 4x2 + y2 = 16 0.5x2 + y2 = 16 4x2 + y2 = 16 0.5x2 + y2 = 16 4x2 + y2 = 16 3.3.8

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**Using a Graphing Calculator [cont’d]**

Using your graphing calculator, graph the following equations: d) x2 + 4y2 = 16 e) x y2 = 16 f) x2 + Cy2 = 16, when C = 0 x2 + Cy2 = 16, when C = 0 x2 + y2 = 16 x2 + y2 = 16 x2 + y2 = 16 x2 + 4y2 = 16 x2 + 4y2 = 16 x2 + 4y2 = 16 x y2 = 16 x y2 = 16 3.3.9

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**Using a Graphing Calculator [cont’d]**

Conclusions: 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. 3.3.10

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**Assignment Suggested Questions: Pages 141 and 142 A 1-25 odd, 27-32**

B , 49, 50, 51, 54, 56, 58 (graph), 62 3.3.11

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