Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 347: 24


$(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=4$ Center: $(\frac{1}{4},\frac{1}{4})$ Radius: $r=2$

Work Step by Step

$2x^2+2y^2-2x-2y-7=0$ (divide the equation by 2): $x^2+y^2-x-y-\frac{7}{2}=0$ Notice that $-x=-2(\frac{1}{2})x$ and that $(\frac{1}{2})^2=\frac{1}{4}$ Notice that $-y=-2(\frac{1}{2})y$ and that $(\frac{1}{2})^2=\frac{1}{4}$ $x^2-x+\frac{1}{4}-\frac{1}{4}+y^2-y+\frac{1}{4}-\frac{1}{4}-\frac{7}{2}=0$ $(x^2-x+\frac{1}{4})+(y^2-y+\frac{1}{4})=\frac{7}{2}+\frac{1}{4}+\frac{1}{4}$ $(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=\frac{7}{2}+\frac{2}{4}=\frac{7}{2}+\frac{1}{2}=4$ $(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=4$ The equation of a circle in standard form: $(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius $(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=2^2$ Center: $(\frac{1}{4},\frac{1}{4})$ Radius: $r=2$
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