College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Cumulative Review: 10

Answer

center: $(-2, 1)$ radius = $3$ units Refer to the image below for the graph.
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Work Step by Step

Add 4 to both sides of the equation to obtain: $x^2+4x+y^2-2y=4$ Group the terms with the same variables to obtain: $(x^2+4x)+(y^2-2y)=4$ Complete the square for each group by adding $(\frac{4}{2})^2=2^2=4$ in the x-group and adding $(\frac{-2}{2})^2=(-1)^2=1$ to the y-group. Add 4 and 1 to the right side to maintain the equality of both sides. $(x^2+4x+4) + (y^2-2y+1) = 4+4+ 1 \\(x^2+4x+4) + (y^2-2y+1) = 9$ Factor each trinomial to obtain: $(x+2)^2 + (y-1)^2=9 \\(x+2)^2+(y-1)^2=3^2$ RECALL: The standard form of a circle's equation is $(x-h)^2 + (y-k)^2=r^2$ where $(h, k)$ is its center and $r$ = radius. Thus, the circle $(x+2)^2+(y-1)^2=3^2$ has its center at $(-2, 1)$ and its radius is $3$ units. To graph the circle, perform the following steps; (1) Plot the center $(-2, 1)$. (2) With a radius of $3$, plot the points that are 3 units directly above, below, to the left, and to the right of the circle (3) Connect the four points (not including the center of the circle) using a smooth curve to form a circle. (Refer to the attached image in the answer part above for the graph.)
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