Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 646: 75


center $(2,0,-1)$, radius $r=\frac{3\sqrt 2}{2}$

Work Step by Step

$2x^2+2y^2+2z^2-8x+4z=-1$, $2(x^2-4x+4)+2(y)^2+2(z^2+2z+1)=-1+8+2$, $2(x-2)^2+2(y)^2+2(z+1)^2=9$, $(x-2)^2+(y)^2+(z+1)^2=\frac{9}{2}$, Thus center $(2,0,-1)$, radius $r=\frac{3\sqrt 2}{2}$
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.