Precalc0011to0016-page21

Precalc0011to0016-page21 - in both groups. x 2 4 x + 4 ( )...

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(Section 0.13: The Cartesian Plane and Circles) 0.13.6 Example 3 (Finding the Standard Form of the Equation of a Circle) A circle has as its equation: 4 x 2 + 4 y 2 ± 16 x + 4 y ± 11 = 0 Find the standard form of this equation, and identify the center and the radius of the circle. § Solution The common coefficient of x 2 and y 2 is 4, so we will divide both sides of the equation by 4. 4 x 2 + 4 y 2 ± 16 x + 4 y ± 11 = 0 x 2 + y 2 ± 4 x + y ± 11 4 = 0 Now, we group together the x 2 and x terms, and we group together the y 2 and y terms. We isolate constant terms on the right-hand side. x 2 ± 4 x () + y 2 + y () = 11 4 We now Complete the Square (CTS)
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Unformatted text preview: in both groups. x 2 4 x + 4 ( ) + y 2 + y + 1 4 = 11 4 + 4 + 1 4 WARNING 2 : Do not forget to add 4 and 1 4 to the right-hand side, also. We now factor both of the resulting Perfect Square Trinomials (PSTs). x 2 ( ) 2 + y + 1 2 2 = 7 We now have the desired form, although the equation could be rewritten as: x 2 ( ) 2 + y 1 2 2 = 7 The circle has center 2, 1 2 and radius 7 ....
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This document was uploaded on 12/29/2011.

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