College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.8 - Page 319: 36

Answer

$(x+3)^{2}+(y-5)^{2}=9$

Work Step by Step

We know that the standard form of the equation of a circle with center $(h,k)$ and radius $r$ is $(x-h)^{2}+(y-k)^{2}=r^{2}$. Therefore, we can plug in the given values for $h$, $k$, and $r$ and simplify. $(x-(-3))^{2}+(y-5)^{2}=(3)^{2}$ $(x+3)^{2}+(y-5)^{2}=9$
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