College Algebra (11th Edition)

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

Chapter 2 - Section 2.2 - Circles - 2.2 Exercises - Page 187: 46

Answer

$(x+4)^2+(y-3)^2=106$

Work Step by Step

A circle has the form: $(x-h)^2+(y-k)^2=r^2$ with center $(h,k)$ and radius $r$ We are told that our circle has center $(-4,3)$ (thus $h=-4,k=3$) and passes through the point $(5,8)$. So: $(x+4)^2+(y-3)^2=r^2$ We plug in $x=5,y=8$: $(5+4)^2+(8-3)^2=r^2$ $9^2+5^2=r^2$ $81+25=r^2$ $106=r^2$ $r=\sqrt{106}$ Thus the equation is: $(x+4)^2+(y-3)^2=106$
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.