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: 69

Answer

See graph. $(0,-3)$ and $(2,-1)$. See explanations.
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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 $(2-2)^2+(-1+3)^2=4, (-1)=(2)-3$
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