Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - Cumulative Review - Page 678: 37

Answer

See below

Work Step by Step

Given $x^2+y^2+12x-4y+15=0$ Rewrite as:$(x^2+12x+36)-36+(y^2-4y+4)-4+15=0\\(x+6)^2+(y-2)^2=25$ Compare the given equation to the standard form of an equation of a circle. You can see that the graph is a circle with center at $(h, k) =(-6, 2)$ and radius $r=5$.
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