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: 8

Answer

center,$(h,k)=(-\frac{1}{4},-1)$ and $r= 1$

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: $16x^{2}+16 y^{2}+8x +32y+1= 0$ The above equation can be written in the standard equation of the circle as follows: $(x-(-\frac{1}{4}))^{2}+ (y-(-1))^{2} =1^{2}$ Hence, $(h,k)=(-\frac{1}{4},-1)$ and $r= 1$
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