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: 30


A circle with center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$

Work Step by Step

Consider $r=3\sqrt 3\sin \theta$ or, $r^2=(3\sqrt 3) r \sin \theta$ This implies that $x^2+y^2-3\sqrt 3y=0$ This can be re-arranged as: $x^2+(y-\dfrac{3\sqrt 3}{2})^2=\dfrac{27}{4}$ Hence, we get the equation of a circle with center $(0,\dfrac{3\sqrt 3}{2})$ and radius $\dfrac{3\sqrt 3}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.