Functions Modeling Change: A Preparation for Calculus, 5th Edition

Published by Wiley
ISBN 10: 1118583191
ISBN 13: 978-1-11858-319-7

Chapter 8 - Triangle Trigonometry and Polar Coordinates - 8.3 Polar Coordinates - Exercises and Problems for Section 8.3 - Exercises and Problems - Page 348: 24


$$ x^2+y^2=6x$$

Work Step by Step

We know the following: $$r=\sqrt{x^2+y^2} \\ cos\theta = \frac{x}{\sqrt{x^2+y^2} } $$ Thus: $$\sqrt{x^2+y^2} = \frac{6x}{\sqrt{x^2+y^2} } \\ x^2+y^2=6x$$
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