Precalculus (6th Edition)

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

Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises: 9

Answer

$\\\color{blue}{(x+8)^2+(y-1)^2=289}$

Work Step by Step

RECALL: The center-radius form of a circle whose center is at $(h, k)$ and whose radius is $r$ units is: $(x-h)^2+(y-k)^2=r^2$ The given circle has its center at $(-8, 1)$ Thus, the tentative equation of the circle is : $(x-(-8))^2+(y-1)^2=r^2 \\(x+8)^2+(y-1)^2=r^2$ To find the value of $r^2$, substitute the x and y values of the given point on the circle $(0, 16)$ to obtain: $(x+8)^2+(y-1)^2=r^2 \\(0+8)^2+(16-1)^2=r^2 \\8^2+15^2=r^2 \\64+225=r^2 \\289=r^2$ Therefore, the equation of the circle is: $\\\color{blue}{(x+8)^2+(y-1)^2=289}$
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