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


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

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.
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