College Algebra (10th Edition)

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

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding - Page 186: 29

Answer

(a) The circle has a center (-2,2) and a radius of 3 units. (b) See below. (c) There are two x-intercepts: $(-4.24,0) \text{ and } (0.24,0)$. There are two y-intercepts: $(0,-0.24) \text{ and } (0,4.24)$.

Work Step by Step

First, we need to make sure that the equation is in standard form $(x-h)^2+(y-k)^2=r^2$ In this case, we need to complete the square to achieve the standard form: $x^2+y^2+4x-4y-1=0$ $x^2+4x+y^2-4y=1$ $x^2+4x+(\frac{4}{2})^2+y^2-4y+(\frac{4}{2})^2=1+(\frac{4}{2})^2+(\frac{4}{2})^2$ $(x+2)^2+(y-2)^2=1+4+4$ $(x+2)^2+(y-2)^2=9$ $(x+2)^2+(y-2)^2=3^2$ Now we can find the center (h,k) and the radius. The x-intercepts are all points of a graph when y=0: $(x+2)^2+(0-2)^2=9$ $(x+2)^2+4=9$ $(x+2)^2=5$ $\sqrt{(x+2)^2}=\sqrt5$ There are two x-intercepts: $x_1+2=-\sqrt5\rightarrow x_1=-\sqrt5-2\approx-4.24$ $x_2+2=\sqrt5\rightarrow x_2=\sqrt5-2\approx0.24$ The y-intercepts are all points of a graph when x=0: $(0+2)^2+(y-2)^2=9$ $4+(y-2)^2=9$ $(y-2)^2=5$ $\sqrt{(y-2)^2}=\sqrt{5}$ There are two y-intercepts: $y_1-2=-\sqrt5\rightarrow y_1=-\sqrt5+2\approx-0.24$ $y_2-2=\sqrt5\rightarrow y_2=\sqrt5+2\approx4.24$
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