Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.1 - Graphs of Equations - 1.1 Exercises - Page 79: 65

Answer

$(x+3)^2+(y+3)^2=61$

Work Step by Step

The middlepoint between $(3,2)$ and $(-9,-8)$ is the center of the circle: $\frac{(3,2)+(-9,-8)}{2}=(-3,-3)$, that is: $h=-3$, $k=-3$ The distance from the center to an endpoint is the radius: $r=\sqrt {(-3-3)^2+(-3-2)^2}=\sqrt {36+25}=\sqrt {61}$ Equation of a circle: $(x−h)^2+(y−k)^2=r^2$ (standard form) $[x-(-3)]^2+[y-(-3)]^2=(\sqrt {61})^2$ $(x+3)^2+(y+3)^2=61$
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