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: 41


$(x-1)^2+(y+3)^2+(z+4)^2=25$ Center: $(1,-3,-4)$ Radius: $5$

Work Step by Step

$x^2+y^2+z^2-2x+6y+8z+1=0$ $x^2+y^2+z^2-2x+6y+8z+1=(x-1)^2-1+(y+3)^2-9+(z+4)^2-16+1$ $=(x-1)^2+(y+3)^2+(z+4)^2-25=0$ $(x-1)^2+(y+3)^2+(z+4)^2=25$ Center: $(1,-3,-4)$ Radius: $5$
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