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: 20

Answer

$\color{blue}{(x+3)^2+(y+2)^2=36}$ Refer to the graph below.
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Work Step by Step

RECALL: The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units. Thus, the given circle's equation in center-radius form is: $(x-(-3))^2+(y-(-2))^2=6^2 \\\color{blue}{(x+3)^2+(y+2)^2=36}$ To graph the circle, perform the following steps: (1) Plot the center. (2) Plot the points 6 units above, below, to the left, and to the right of the center. (3) Connect the four points in Step 2 using a curve to form a circle. (Refer to the attached image in the answer part above for the graph.)
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