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: 31


A circle with center $(\sqrt 2,0)$ and radius $\sqrt 2$

Work Step by Step

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