Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Review Exercises - Page 798: 44


$(x-1.5)^2+y^2=2.25$ See graph.

Work Step by Step

Step 1. Multiply the equation with $r$; we have $r^2=3r\ cos\theta$. Step 2. Using $r^2=x^2+y^2$ and $r\ cos\theta=x$, we have $x^2+y^2=3x$ which gives $x^2+y^2-3x=0$ and $(x-1.5)^2+y^2=1.5^2$ or $(x-1.5)^2+y^2=2.25$ Step 3. We can identify the above equation as a circle with center $(1.5,0)$ and radius $r=1.5$ Step 4. See graph.
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