Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises: 135

Answer

$(x+4)^2+(y-3)^2=4$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Center-Radius Form of the equation of circles to get the equation with the given center, $C( -4,3 ),$ and the given radius, $r= 2 .$ $\bf{\text{Solution Details:}}$ With the given center, then $h= -4 $ and $k= 3 .$ Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2, $ the equation of the circle is \begin{array}{l}\require{cancel} (x-(-4))^2+(y-3)^2=(2)^2 \\\\ (x+4)^2+(y-3)^2=4 .\end{array}
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