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: 58


Center is $(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius is $\dfrac{\sqrt {29}}{3}$

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 $r$ Given: $3x^2+3y^2+3z^2+2y-2z=9$ or, $2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$ $ \implies x^2+(y+\dfrac{1}{3})^2+(z-\dfrac{1}{3})^2=\dfrac{29}{9}$ $ \implies (x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\sqrt{\dfrac{29}{9}})^2$ and $(x-0)^2+(y -(-\dfrac{1}{3}))^2+(z-\dfrac{1}{3})^2=(\dfrac{\sqrt {29}}{3})^2$ Thus, we have Center is $(0,-\dfrac{1}{3},\dfrac{1}{3})$ and radius is $\dfrac{\sqrt {29}}{3}$
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