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: 41



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 $( 0,0 )$ is \begin{array}{l}\require{cancel} (x-0)^2+(y-0)^2=r^2 \\\\ x^2+y^2=r^2 .\end{array} Since the given point, $( -3,4 ),$ is on the circle, then substitute these coordinates into the $x$ and $y$ variables, respectively, of the equation above. That is, \begin{array}{l}\require{cancel} (-3)^2+(4)^2=r^2 \\\\ 9+16=r^2 \\\\ r^2=25 .\end{array} Using the given center and the solved radius, the equation of the circle is $ x^2+y^2=25 .$
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