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 - Page 854: 43



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 $( -4,1 )$ is \begin{array}{l}\require{cancel} (x-(-4))^2+(y-1)^2=r^2 \\\\ (x+4)^2+(y-1)^2=r^2 .\end{array} Since the given point, $( -2,5 ),$ is on the circle, then substitute these coordinates into the $x$ and $y$ variables, respectively, of the equation above. \begin{array}{l}\require{cancel} (-2+4)^2+(5-1)^2=r^2 \\\\ (2)^2+(4)^2=r^2 \\\\ 4+16=r^2 \\\\ r^2=20 .\end{array} Using the given center and the solved radius, the equation of the circle is $ (x+4)^2+(y-1)^2=20 .$
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