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: 23


$(x-1)^2+(y+3)^2=1$ Center: $(1,-3)$ Radius: $r=1$

Work Step by Step

$x^2+y^2-2x+6y+9=0$ Notice that $-2x=-2(1)x$ and that $1^2=1$ Notice that $6y=2(3)y$ and that $3^2=9$ $x^2-2x+1-1+y^2+6y+9-9+9=0$ $(x^2-2x+1)+(y^2+6y+9)=9-9+1$ $(x-1)^2+(y+3)^2=1$ 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 $(x-1)^2+[y-(-3)]^2=1^2$ Center: $(1,-3)$ Radius: $r=1$
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