## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 12: Vectors and the Geometry of Space - Section 12.1 - Three-Dimensional Coordinate Systems - Exercises 12.1 - Page 696: 30

#### Answer

(a) $(x-3)^2+(y-4)^2=1 ,z=1$ (b) $(y-4)^2+(z-1)^2=1,x=-3$ (c) $(x-3)^2+(z-1)^2=1, y=4$

#### Work Step by Step

Let us consider $r$ as the radius of a circle and $(x_0,y_0,z_0)$ as center. Points to be noted: i) Equation of a circle lies in a plane parallel to $xy$ plane is represented as: $(x-x_0)^2+(y-y_0)^2=r^2$; $z=z_0$ ii) Equation of a circle lies in a plane parallel to $yz$ plane is represented as:$(y-y_0)^2+(z-z_0)^2=r^2$; $x=x_0$ iii) Equation of a circle lies in a plane parallel to $xy$ plane is represented as: $(x-x_0)^2+(z-z_0)^2=r^2$; $y=y_0$ Hence, our equations are: (a) $(x-3)^2+(y-4)^2=1 ,z=1$ (b) $(y-4)^2+(z-1)^2=1,x=-3$ (c) $(x-3)^2+(z-1)^2=1, y=4$

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