## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises - Page 296: 10

#### Answer

$\color{blue}{(x-3)^2+(y+6)^2=36}$

#### 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 $(3, -6)$ Thus, the tentative equation of the circle is : $(x-3)^2+(y-(-6))^2=r^2 \\(x-3)^2+(y+6)^2=r^2$ Since the circle is tangent to the x-axis, then the point on the x-axis directly above the center $(3, -6)$ must be a point on the circle. This point is $(3, 0)$. To find the value of $r^2$, substitute the x and y values of the point $(3, 0)$ to obtain: $(x-3)^2+(y+6)^2=r^2 \\(3-3)^2+(0+6)^2=r^2 \\0^2+6^2=r^2 \\0+36=r^2 \\36=r^2$ Therefore, the equation of the circle is: $\\\color{blue}{(x-3)^2+(y+6)^2=36}$

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