## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - Chapter 2 Test Prep - Review Exercises - Page 296: 11

#### Answer

$\color{blue}{x^2+y^2=34}$

#### 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 $(0, 0)$ Thus, the tentative equation of the circle is : $(x-0)^2+(y-0)^2=r^2 \\x^2+y^2=r^2$ To find the value of $r^2$, substitute the x and y values of the point $(3, 5)$ to obtain: $3^2+5^2=r^2 \\9+25=r^2 \\34=r^2$ Therefore, the equation of the circle is: $\\\color{blue}{x^2+y^2=34}$

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