University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 599: 11


Circle with equation $x^2+y^2=16$, in the $xy$ plane.

Work Step by Step

Here, the the two equations represent the intersection between the equation of a sphere $x^2+y^2+(z+3)^2=25$ and the plane $z=0$. Now, $x^2+y^2+(0+3)^2=25 \implies x^2+y^2=25-9$ or, $x^2+y^2=16$ The set of points that satisfies the equation $x^2+y^2=16$ is a circle lying in $z=0$, centered at (0,0,0), having a circle of radius $4$ .
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