Precalculus (6th Edition)

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

Chapter 2 - Graphs and Functions - Summary Exercises on Graphs, Circles, Functions, and Equations - Exercises - Page 248: 11

Answer

$(x-2)^2+(y+1)^2=9$ Refer to the graph below.
1517065715

Work Step by Step

RECALL: A circle whose equation is of the form $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and has a radius of $r$. The given circle has its center $(2, -1)$ and a radius of $3$ units. Substituting these values into the equation above gives: $(x-2)^2+(y-(-1))^2=3^2 \\(x-2)^2+(y+1)^2=9$ Refer to the graph in the answer part above.
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.