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: 56


Center is $(0,3,-4 )$ and radius is $5$

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: $x^2+y^2+z^2-6y+8z=0$ This can be written as: $x^2+y^2-6y+z^2+8z=0 \implies x^2+(y-3)^2+(z+4)^2=0+9+16$ or, $x^2+(y-3)^2+(z+4)^2=25 \implies (x -0)^2+(y-3)^2+(z-(-4))^2=5$ Thus, we have Center is $(0,3,-4 )$ and radius is $5$
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