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: 39



Work Step by Step

Endpoints of the diameter are $(2,0,0)$ and $(0,6,0)$. The center of the sphere is therefore the midpoint of these two points. Center=$\left(\frac{2+0}{2},\frac{0+6}{2},\frac{0+0}{2}\right)=(1,3,0)$. The radius is $\frac{1}{2}\times$distance between the two points. $r=\frac{1}{2}\sqrt{(2-0)^2+(0-6)^2+(0-0)^2}=\frac{1}{2}\sqrt{40}=\frac{1}{2}\times2\sqrt{10}=\sqrt{10}$ The equation of the sphere is therefore $(x-1)^2+(y-3)^2+(z-0)^2=\sqrt{10}^2$ =$$(x-1)^2+(y-3)^2+z^2=10$$
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