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: 33

Answer

(a) The circle has a center (3,-2) and a radius of 5 units. (b) See below. (c) There are two x-intercepts: $(-1.58,0) \text{ and } (7.58,0)$. There are two y-intercepts: $(0,-6) \text{ and } (0,2)$.

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: $2x^2+2y^2-12x+8y-24=0$ $2(x^2+y^2-6x+4y-12)=0$ $x^2+y^2-6x+4y-12=0$ $x^2-6x+y^2+4y=12$ $x^2-6x+(\frac{6}{2})^2+y^2+4y+(\frac{4}{2})^2=12+(\frac{6}{2})^2+(\frac{4}{2})^2$ $(x-3)^2+(y+2)^2=12+9+4$ $(x-3)^2+(y+2)^2=25$ $(x-3)^2+(y+2)^2=5^2$ Now we can find the center (h,k) and the radius. The x-intercepts are all points of a graph when y=0: $(x-3)^2+(0+2)^2=25$ $(x-3)^2+4=25$ $(x-3)^2=21$ $\sqrt{(x-3)^2}=\sqrt{21}$ There are two x-intercepts: $x_1-3=-\sqrt{21}\rightarrow x_1=-\sqrt{21}+3\approx-1.58$ $x_2-3=\sqrt{21}\rightarrow x_2=\sqrt{21}+3\approx7.58$ The y-intercepts are all points of a graph when x=0: $(0-3)^2+(y+2)^2=25$ $9+(y+2)^2=25$ $(y+2)^2=16$ $\sqrt{(y+2)^2}=\sqrt{16}$ There are two y-intercepts: $y_1+2=-4\rightarrow y_1=-6$ $y_2+2=4\rightarrow y_2=2$
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