Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises - Page 296: 9



Work Step by Step

RECALL: The center-radius form of a circle whose center is at $(h, k)$ and whose radius is $r$ units is: $(x-h)^2+(y-k)^2=r^2$ The given circle has its center at $(-8, 1)$ Thus, the tentative equation of the circle is : $(x-(-8))^2+(y-1)^2=r^2 \\(x+8)^2+(y-1)^2=r^2$ To find the value of $r^2$, substitute the x and y values of the given point on the circle $(0, 16)$ to obtain: $(x+8)^2+(y-1)^2=r^2 \\(0+8)^2+(16-1)^2=r^2 \\8^2+15^2=r^2 \\64+225=r^2 \\289=r^2$ Therefore, the equation of the circle is: $\\\color{blue}{(x+8)^2+(y-1)^2=289}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.