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: 28

Answer

(a) The circle has a center (-2,-1) and a radius of 5 units. (b) See below. (c) There are two x-intercepts: $(-6.9,0) \text{ and } (2.9,0)$. There are two y-intercepts: $(0,-5.58) \text{ and } (0,3.58)$.

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: $x^2+y^2+4x+2y-20=0$ $x^2+4x+y^2+2y=20$ $x^2+4x+(\frac{4}{2})^2+y^2+2y+(\frac{2}{2})^2=20+(\frac{4}{2})^2+(\frac{2}{2})^2$ $(x+2)^2+(y+1)^2=20+4+1$ $(x+2)^2+(y+1)^2=25$ $(x+2)^2+(y+1)^2=5^2$ Now we can find the center (h,k) and the radius. The x-intercepts are all points of a graph when y=0: $(x+2)^2+(0+1)^2=25$ $(x+2)^2+1=25$ $(x+2)^2=24$ $\sqrt{(x+2)^2}=\sqrt{24}$ There are two x-intercepts: $x_1+2=-\sqrt{24}\rightarrow x_1=-\sqrt{24}-2\approx-6.9$ $x_2+2=\sqrt{24}\rightarrow x_2=\sqrt{24}-2\approx2.9$ The y-intercepts are all points of a graph when x=0: $(0+2)^2+(y+1)^2=25$ $4+(y+1)^2=25$ $(y+1)^2=21$ $\sqrt{(y+1)^2}=\sqrt{21}$ There are two y-intercepts: $y_1+1=-\sqrt{21}\rightarrow y_1=-\sqrt{21}-1\approx-5.58$ $y_2+1=\sqrt{21}\rightarrow y_2=\sqrt{21}-1\approx3.58$
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