Answer
center $(-1,1,0)$, radius $r=\sqrt 3$
Work Step by Step
$3x^2+3y^2+3z^2+6x-6y=3$,
$3(x^2+2x+1)+3(y^2-2y+1)+3z^2=3+3+3$,
$3(x+1)^2+3(y-1)^2+3(z)^2=9$,
$(x+1)^2+(y-1)^2+(z)^2=3$,
Thus center $(-1,1,0)$, radius $r=\sqrt 3$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
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: 76
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