Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set: 39

Answer

$(x+5)^2+(y+8)^2=300$

Work Step by Step

Using $(x-h)^2+(y-k)^2=r^2$ or the Center-Radius form of the equation of circles, the equation of the circle with center $( -5,-8 )$ and radius, $r= 10\sqrt{3} ,$ is \begin{array}{l}\require{cancel} (x-(-5))^2+(y-(-8))^2=\left( 10\sqrt{3} \right)^2 \\\\ (x+5)^2+(y+8)^2=100\cdot3 \\\\ (x+5)^2+(y+8)^2=300 .\end{array}
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