Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 11 - Vectors and the Geometry of Space - 11.2 Exercises - Page 763: 44

Answer

$(x-3)^2+(y-\frac{1}{2})^2+(z+1)^2=16$ Center: $(3,\frac{1}{2},-1)$ and Radius: $4$

Work Step by Step

$4x^2+4y^2+4z^2-24x-4y+8z-23=0$, and so dividing by $4$ $x^2+y^2+z^2-6x-y+2z-\frac{23}{4}=0$ $=(x-3)^2-9+(y-\frac{1}{2})^2-\frac{1}{4}+(z+1)^2-1-\frac{23}{4}$ $=(x-3)^2+(y-\frac{1}{2})^2+(z+1)^2-16=0$ $(x-3)^2+(y-\frac{1}{2})^2+(z+1)^2=16$ Center: $(3,\frac{1}{2},-1)$ and Radius: $4$
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