Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 32

Answer

A circle with center $(-3,0)$ and radius $3$

Work Step by Step

Consider $r=-6\cos \theta$ or, $r^2=-6r \cos \theta$ This implies that $x^2+y^2=-6x$ This can be re-arranged as: $(x+3)^2+y^2=9$ Hence, we get the equation of a circle with center $(-3,0)$ and radius $3$
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