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: 12

Answer

$x^2+(y-2)^2=4$ Refer to the graph below.
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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 at $(0, 2)$ and tangent to the x-axis. Since the center is $2$ units away from the x-axis, then the circle has a radius of $2$ units. Therefore, the equation of the circle is: $(x-0)^2+(y-2)^2=2^2 \\x^2+(y-2)^2=4$ Refer to the graph in the answer part above.
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