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: 34


$x^{2}+y^{2}=3,\quad z=0$

Work Step by Step

(distance from $(0,0,1)$ to (x,y,z))=2 $\sqrt{x^{2}+y^{2}+(z-1)^{2}}=2$ (distance from $(0,0,-1)$ to (x,y,z))=2 $\sqrt{x^{2}+y^{2}+(z+1)^{2}}=2$ 2=2, so we equate the LHS's $\sqrt{x^{2}+y^{2}+(z-1)^{2}}=\sqrt{x^{2}+y^{2}+(z+1)^{2}}\quad$ square both sides $ x^{2}+y^{2}+(z-1)^{2}=x^{2}+y^{2}+(z+1)^{2}\quad$ ... simplify $(z-1)^{2}=(z+1)^{2}$ $z^{2}-2z+1=z^{2}+2z+1$ $-4z=0$ $z=0$ (this set of points is in the xy-plane) Substitute $z=0$ in $x^{2}+y^{2}+(z-1)^{2}=4,$ $x^{2}+y^{2}+1=4,\quad z=0$ $x^{2}+y^{2}=3,\quad z=0$ (circle of radius $\sqrt{3}$ in the xy-plane, centered at the origin).
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