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: 14

Answer

$\color{blue}{(x-5)^2+(y-6)^2=10}$

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 $(5, 6)$ Thus, the tentative equation of the circle is : $(x-5)^2+(y-6)^2=r^2$ To find the value of $r^2$, substitute the x and y values of the point $(4, 9)$ to obtain: $(4-5)^2+(9-6)^2=r^2 \\(-1)^2+3^2=r^2 \\1+9=r^2 \\10=r^2$ Therefore, the equation of the circle is: $\\\color{blue}{(x-5)^2+(y-6)^2=10}$
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