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: 26

Answer

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

Gather terms containing x and y in separate parentheses $(x^{2}-4x)+(y^{2}+4y)=0$ Complete squares: $x^{2}-4x+4-4=(x-2)^{2}-4$ $y^{2}+4y+4-4=(y+2)^{2}-4$ The equation becomes $(x-2)^{2}-4+(y+2)^{2}-4=0$ $(x-2)^{2}+(y+2)^{2}=8$ A circle with center at $C(2,-2)$ and radius $2\sqrt{2}$ x-intercepts $:\quad$ $\left[\begin{array}{l} (x-2)^{2}+(0+2)^{2}=8\\ (x-2)^{2}+4=8\\ (x-2)^{2}=4\\ x-2=\pm 2 \\ x=2\pm 2\\ (0,0),\quad(4,0) \end{array}\right] $ y-intercepts $:\quad \left[\begin{array}{l} (0-2)^{2}+(y+2)^{2}=8\\ 4+(y+2)^{2}=8\\ (y+2)^{2}=4\\ y+2=\pm 2\\ y=-2\pm 2\\ y=0,\quad y=-4\\ (0,-4),\quad(0,0) \end{array}\right] $
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