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: 73


$ \ Radius = 3$ and Center at $(2,-2,-1)$

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 $x^2+y^2+z^2-4x+4y+2z=0$ We complete the square as follows: $(x^2-4x) +(y^2+4y) + (z^2 +2z) =0 \\ (x^2-4x+4) +(y^2 +4y+4) +(z^2 +2z+1) =0+4+4+1 \\ (x -2)^2+(y +2)^2+(z+1)^2 = 9$ On comparing the above equation with equation (1), we have: $ \ Radius, r= \sqrt 9=3$ and Center at $(2,-2,-1)$
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