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: 56

Answer

$(x-2)^2+(y-6)^2=49$ $C(2,6), r=7$ See graph.

Work Step by Step

Step 1. From the given equation $x^2+y^2-4x-12y-9=0$, we have $(x^2-4x+4)+(y^2-12y+36)=4+36+9$ and $(x-2)^2+(y-6)^2=49$ Step 2. We can identify the center and radius of the circle as $C(2,6), r=7$ Step 3. See graph.
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