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: 2

Answer

$\color{blue}{(x-3)^2+(y-6)^2=16}$

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. Thus, the equation of a circle with center at $(3, 6)$ and a radius of $4$ is: $(x-3)^2+y-6)^2=4^2 \\\color{blue}{(x-3)^2+(y-6)^2=16}$
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