## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises: 8

#### Answer

$\color{blue}{(x-\sqrt5)^2+(y+\sqrt7)^2=3}$

#### 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 $(\sqrt5, -\sqrt7)$ and has a radius of 15 units. Thus, its equation in center-radius form is : $(x-\sqrt5)^2+(y-(-\sqrt7))^2=(\sqrt3)^2 \\\color{blue}{(x-\sqrt5)^2+(y+\sqrt7)^2=3}$

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