Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 10 - Section 10.1 - The Parabola and the Circle - Exercise Set - Page 610: 74

Answer

See graph

Work Step by Step

$$x^2+y^2-8y+5=0$$ $$x^2+y^2-8y=-5$$ Complete the square. $$x^2+y^2-8y+16=-5+16$$ $$x^2+(y-4)^2=11$$ This is a circle with center (0,4) and a radius of length $\sqrt{11}$. Points are: $$(\sqrt{11},4)$$ $$(-\sqrt{11},4)$$ $$(0,4+\sqrt{11})$$ $$(0,4-\sqrt{11})$$ Approximate these values using a calculator, plot on a graph, and connect with a circle.
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