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: 57


Center is $(-\dfrac{1}{4},-\dfrac{1}{4},-\dfrac{1}{4})$ and radius is $\dfrac{5 \sqrt 3}{4}$

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: $2x^2+2y^2+2z^2+x+y+z=9$ $\implies 2(x^2+\dfrac{1}{2}x)+2(y^2+\dfrac{1}{2}y)+2(z^2+\dfrac{1}{2}z)=9$ $ \implies 2(x^2+\dfrac{1}{2}x+\dfrac{1}{16})+2(y^2+\dfrac{1}{2}y+\dfrac{1}{16})+2(z^2+\dfrac{1}{2}z+\dfrac{1}{16})=9+\dfrac{2}{16}+\dfrac{2}{16}+\dfrac{2}{16}$ or, $(x +\dfrac{1}{4})^2+(y +\dfrac{1}{4})^2+(z +\dfrac{1}{4})^2=(\sqrt{\dfrac{75}{16}})^2$ and $(x -(-\dfrac{1}{4}))^2+(y -(-\dfrac{1}{4}))^2+(z -(-\dfrac{1}{4}))^2=(\dfrac{5 \sqrt 3}{4})^2$ Thus, we have Center is $(-\dfrac{1}{4},-\dfrac{1}{4},-\dfrac{1}{4})$ and radius is $\dfrac{5 \sqrt 3}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.