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 695: 10


circle with equation $x^2+z^2=9$ in the plane $y=-4$

Work Step by Step

Since, we have two equations shows the intersection between the equation of sphere $x^2+y^2+z^2=25$ and plane $y=-4$. It can be rewritten as: $x^2+(-4)^2+z^2=25$ or, $ x^2+z^2=25-16$ and $x^2+z^2=9$ Thus, we have the set of points a circle with equation $x^2+z^2=9$ in the plane $y=-4$ , center: $(0,-4,0)$ with circle of radius $3$ .
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