Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 8 - Polar Coordinates; Vectors - Section 8.6 Vectors in Space - 8.6 Assess Your Understanding - Page 646: 74


$ \ Radius=2$ and Center at $(2,0,0)$

Work Step by Step

Let $(a,b,c)$ be the center of the sphere, $(x,y,z)$ be an arbitrary point of the sphere, and $r$ be the distance between $(a,b,c)$ and $(x,y,z)$. Then, using the Pythagorean Theorem, the standard form of the equation of the sphere can be expressed as: $r=\sqrt {(x-a)^2+(y-b)^2+(z-c)^2}$ or, $(x-a)^2+(y-b)^2+(z-c)^2=r^2 ...(1)$ We are given: $x^2+y^2+z^2-4x=0$ We complete the square as follows: $(x^2-4x) +y^2 + z^2 =0 \\ (x^2-4x+4) +y^2 +z^2 =0+4 \\ (x -2)^2+y^2+z^2 = 4$ On comparing the above equation with equation (1), we have: $ \ Radius, r= \sqrt 4=2$ and Center at $(2,0,0)$
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