Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.3 - Lines, Circles, and Parabolas - Exercises A.3 - Page AP-17: 24

Answer

A circle with center at $C(-2,2)$ and radius $2$ .
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Work Step by Step

Gather terms containing x and y in separate parentheses $(x^{2}+4x)+(y^{2}-4y)=-4$ Complete squares: $x^{2}+4x=x^{2}+2\cdot x\cdot 2+2^{2}-2^{2}=(x+2)^{2}-4$ $y^{2}-4y=y^{2}-2\cdot y\cdot 2+2^{2}-2^{2}=(y-2)^{2}-4$ The equation becomes $(x+2)^{2}-4+(y-2)^{2}-4=-4$ $(x+2)^{2}+(y-2)^{2}=4$ A circle with center at $C(-2,2)$ and radius $2$ x-intercepts $:\quad\left[\begin{array}{l} (x+2)^{2}+(0-2)^{2}=4\\ (x+2)^{2}+4=4\\ x=-2\\ (-2,0) \end{array}\right]$, y-intercepts $:\quad\left[\begin{array}{l} (0+2)^{2}+(y-2)^{2}=4\\ 4+(y-2)^{2}=4\\ y=2\\ (0,2) \end{array}\right]$
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