Precalculus (6th Edition)

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

Chapter 2 - Graphs and Functions - 2.2 Circles - 2.2 Exercises - Page 199: 3

Answer

$(4, -7)$

Work Step by Step

RECALL: The circle $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and a radius of $r$ units. The given equation can be written as: $(x-4)^2+(y-(-7))^2=3^2$ Thus, the center of the circle is at $(4, -7)$.
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