Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.1 - Three-Dimensional Coordinate Systems - 12.1 Exercises - Page 797: 22

Answer

$(x-3)^{2}$ $+$ $(y-5)^{2}$ $+$ $(z+3)^{2}$ $=$ $41$

Work Step by Step

Because the diameters has endpoints $($ $5$, $4$, $3$ $)$ and $($ $1$, $6$, $-$$9$ $)$, so the center of the sphere is $($ $\frac{5+1}{2}$, $\frac{4+6}{2}$, $\frac{3+(-9)}{2}$$)$ $=$ $($ $3$, $5$, $-3$ $)$ the radius of the sphere is $r$ $=$ $\sqrt{(5-3)^{2}+(4-5)^{2}+(3-(-3))^{2}}$ $=$ $\sqrt {41}$ Therefore, the equation of the sphere is $(x-3)^{2}$ $+$ $(y-5)^{2}$ $+$ $(z+3)^{2}$ $=$ $41$
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