Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 4 - 4.4 - Translations of Conics - 4.4 Exercises - Page 347: 20

Answer

Center: $(0,-12)$ Radius: $r=2\sqrt 6$

Work Step by Step

The equation of a circle in standard form: $(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius $x^2+(y+12)^2=24$ $(x-0)^2+[y-(-12)]^2=(\sqrt {24})^2$ Center: $(0,-12)$ Radius: $r=\sqrt {24}=2\sqrt 6$
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.