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: 36

Answer

$(x-5)^2+(y-6)^2=11$

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 $( 5,6 )$ and radius, $r= \sqrt{11} ,$ is \begin{array}{l}\require{cancel} (x-5)^2+(y-6)^2=\left( \sqrt{11} \right)^2 \\\\ (x-5)^2+(y-6)^2=11 .\end{array}
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