## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 53

#### Answer

$(x + 1)^2+(y -\dfrac{1}{2})^2+(z+ \dfrac{2}{3})^2=\dfrac{16}{81}$

#### Work Step by Step

We know that the standard equation of a sphere is written as: $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2\ \ \$ (1) Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere. Since, we have $x_0=-1,y_0=\dfrac{1}{2},z_0=\dfrac{-2}{3}, r=\dfrac{4}{9}$ Plug all the above values in equation (1), then we get Thus,, $(x - (-1))^2+(y -(\dfrac{1}{2}))^2+(z - (\dfrac{-2}{3}))^2=(\dfrac{4}{9})^2$ Hence, $(x + 1)^2+(y -\dfrac{1}{2})^2+(z+ \dfrac{2}{3})^2=\dfrac{16}{81}$

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