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 - Section 11.7 - Conics in Polar Coordinates - Exercises 11.7 - Page 686: 59

Answer

$r=10 \sin\theta$ .

Work Step by Step

The circle passes through the origin, as (0,0) satisfies the Cartesian equation. The radius is $5$, and the center is at $(0,5)$. In polar coordinates, the center lies at $r_{0}=5, \theta_{0}=\pi/2,\qquad P(5,\pi/2)$ A circle passing through the origin, of radius $a$, centered at $P_{0}(r_{0}, \theta_{0}),$ has the polar equation $r=2a\cos(\theta-\theta_{0})$ So this circle has the equation $r=2(5)\displaystyle \cos(\theta-\frac{\pi}{2})$ or $r=10\displaystyle \cos(\theta-\frac{\pi}{2})$ Applying the trigonometric identity $\displaystyle \cos(\theta\pm\frac{\pi}{2})=\mp\sin\theta$, we can rewrite the equation as $r=10 \sin\theta$
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