Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.1 - Three-Dimensional Coordinate Systems - Exercises 12.1 - Page 696: 47


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

Work Step by Step

As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ ) where $(a,b,c)$ represents center and radius of the sphere is $r$ Now, $(x+2)^2+y^2+(z-2)^2=8$ $\implies (x-(-2))^2+(y-0)^2+(z-2)^2=(\sqrt 8)^2$ $ \implies (x-(-2))^2+(y-0)^2+(z-2)^2=( 2 \sqrt 2)^2$ Compare this equation with the standard equation of sphere. Thus, Center $(-2,0,2)$ and radius $2 \sqrt 2$
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