College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Review Exercises - Page 275: 9

Answer

$(x+8)^2+(y-1)^2=289$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Center-Radius Form to find the equation of the circle with the following characteristics: \begin{array}{l}\require{cancel} \text{center } (-8,1) ,\text{ passing through } (0,16) .\end{array} $\bf{\text{Solution Details:}}$ With the given center, then $h= -8 $ and $k= 1 .$ Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ then the equation of the circle is \begin{array}{l}\require{cancel} (x-(-8))^2+(y-1)^2=r^2 \\\\ (x+8)^2+(y-1)^2=r^2 \text{ (*)} .\end{array} Since the circle passes through the point $ (0,16) ,$ then substitute the coordinates in the equation. The equation above becomes \begin{array}{l}\require{cancel} (0+8)^2+(16-1)^2=r^2 \\\\ (8)^2+(15)^2=r^2 \\\\ 64+225=r^2 \\\\ 289=r^2 .\end{array} Substituting $r^2$ in (*), the equation of the circle is \begin{array}{l}\require{cancel} (x+8)^2+(y-1)^2=289 .\end{array}
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.