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: 64

Answer

$(x+2)^2+(y+6)^2=25$

Work Step by Step

Center: $(h,k)=(-2,-6)$ Point on the circle: $(1,-10)$ The distance from the center to the solution point is the radius $r$ of the circle. Finding $r$ using the distance formula: $r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt {(-2-1)^2+[-6-(-10)]^2}=\sqrt {9+16}=\sqrt {25}=5$ Using the standard form of the equation of a circle: $$(x−h)^2+(y−k)^2=r^2$$ $$[(x-(-2)]^2+[y-(-6)]^2=5^2$$ Simplifying: $$(x+2)^2+(y+6)^2=25$$
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