University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Practice Exercises - Page 593: 32


Center $(-3,0)$ and radius $3$

Work Step by Step

Here, we have $r=-6\cos \theta$ This implies that $r^2=-6r \cos \theta$ $x^2+y^2=-6x$ or, $(x+3)^2+y^2=9$ Thus, we have a equation of a circle with center $(-3,0)$ and radius $3$.
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