Precalculus (6th Edition) Blitzer

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

Chapter 9 - Review Exercises - Page 1036: 56

Answer

$(x-3)^2+(y-1)^2=4$; see figure.
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Work Step by Step

Step 1. From the given equations, we have $cos(t)=\frac{x-3}{2}$ and $sin(t)=\frac{y-1}{2}$ Step 2. Using the identify $sin^2t+cos^2t=1$, we have $(\frac{x-3}{2})^2+(\frac{y-1}{2})^2=1$ or $(x-3)^2+(y-1)^2=4$, indicating a circle centered at $(3,1)$ with radius $r=2$ Step 3. Given $0\leq t\lt 2\pi$, we can graph the above equation as shown in the figure.
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