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 600: 47


Center :$(-2,0,2)$ and radius: $2 \sqrt 2$

Work Step by Step

We know that the standard equation of a sphere is written as: $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2$ ...(1) Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere. Since, we have $(x+2)^2+y^2+(z-2)^2=8$ or, $(x-(-2))^2+(y-0)^2+(z-2)^2=(\sqrt 8)^2$ or, $(x-(-2))^2+(y-0)^2+(z-2)^2=( 2 \sqrt 2)^2$ Compare the above equation with equation (1): Center :$(-2,0,2)$ and radius: $2 \sqrt 2$
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