Calculus: Early Transcendentals 8th Edition

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

APPENDIX C - Graphs of Second-Degree Equations - C Exercises: 6

Answer

center $(h,k)=(0,-3)$ and $r= \sqrt 7$

Work Step by Step

An equation of the circle with center $(h,k)$ and radius $r$ is given as: $(x-h)^{2}+(y-k)^{2}=r^{2}$ ...(1) Given: $x^{2}+ y^{2}+6y +2 = 0$ The above equation can be written in the standard equation of the circle as follows: $(x-0)^{2}+ (y-(-3))^{2} = (\sqrt 7)^{2}$ Compare it with equation (1) to have Hence $(h,k)=(0,-3)$ and $r= \sqrt 7$
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