## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 280: 69

#### Answer

See graph. $(0,-3)$ and $(2,-1)$. See explanations.

#### Work Step by Step

Step 1. See graph. Step 2. We can find the intersections as $(0,-3)$ and $(2,-1)$. Step 3. With the original equations, plug-in the coordinates; we have $(0-2)^2+(-3+3)^2=4, (-3)=(0)-3$ and $(2-2)^2+(-1+3)^2=4, (-1)=(2)-3$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.