Answer
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 $3$.
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 10 - Practice Exercises - Page 593: 32
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