Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 280: 70

Answer

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

Step 1. See graph. Step 2. We can find the intersections as $(0,-1)$ and $(3,2)$. Step 3. With the original equations, plug-in the coordinates; we have $(0-3)^2+(-1+1)^2=9, (-1)=(0)-1$ and $(3-3)^2+(2+1)^2=9, (2)=(3)-1$
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