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: 48


Center $(1,\dfrac{-1}{2},-3)$ and radius is $5$

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, $(x -1)^2+(y+\dfrac{1}{2})^2+(z +3)^2=25$ $ \implies (x -1)^2+(y-(-\dfrac{1}{2}))^2+(z -(-3))^2=5^2$ Compare this equation with the standard equation of sphere. Thus, Center $(1,\dfrac{-1}{2},-3)$ and radius is $5$
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