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: 30


Center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$

Work Step by Step

Here, we have $r=3\sqrt 3\sin \theta$ This implies that $r^2=3\sqrt 3r \sin \theta$ $x^2+y^2-3\sqrt 3y=0$ or, $x^2+(y-\dfrac{3\sqrt 3}{2})^2=\dfrac{27}{4}$ Thus, we have a equation of a circle with center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
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