Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.3 - Polar Coordinates - 10.3 Exercises - Page 666: 17

Answer

This is a cirlce with radius$\frac{5}{2}$ and center at $(\frac{5}{2},0)$.

Work Step by Step

$r=5cos\theta$ $r^{2}=5rcos\theta$ To convert polar equation to Cartesian coordinates, we use the equations $rcos \theta=x$ , $rsin \theta=y$ and $x^{2}+y^{2}=r^{2}$ Thus, $x^{2}+y^{2}=5x$ $(x^{2}-5x)+y^{2}=0$ Add $(\frac{5}{2})^{2}$ on both sides. $(x^{2}-2 \cdot \frac{5}{2}x+(\frac{5}{2})^{2})+y^{2}=(\frac{5}{2})^{2}$ $(x-(\frac{5}{2}))^{2}+y^{2}=(\frac{5}{2})^{2}$ This is a cirlce with radius$\frac{5}{2}$ and center at $(\frac{5}{2},0)$.
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