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: 58

Answer

$(x+6)^2+(y-3)^2=49$ $C(-6,3), r=7$ See graph.
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Work Step by Step

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