College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.8 - Page 319: 57

Answer

Center: $(-4,1)$ $;$ Radius: $5$
1508289203

Work Step by Step

$x^{2}+y^{2}+8x-2y-8=0$ Take $8$ to the right side: $x^{2}+y^{2}+8x-2y=8$ Group the terms with $x$ together and the terms with $y$ together: $(x^{2}+8x)+(y^{2}-2y)=8$ Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ inside each parentheses and also on the right side of the equation. For the first group, $b=8$ and for the second group, $b=-2$ $\Big[x^{2}+8x+\Big(\dfrac{8}{2}\Big)^{2}\Big]+\Big[y^{2}-2y+\Big(-\dfrac{2}{2}\Big)^{2}\Big]=8+\Big(\dfrac{8}{2}\Big)^{2}+\Big(-\dfrac{2}{2}\Big)^{2}$ $(x^{2}+8x+16)+(y^{2}-2y+1)=8+16+1$ $(x^{2}+8x+16)+(y^{2}-2y+1)=25$ Factor the expressions inside each parentheses, which are perfect square trinomials: $(x+4)^{2}+(y-1)^{2}=25$ This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$ It can be seen from the obtained equation that $(h,k)=(-4,1)$, so the center of the circle is the point $(-4,1)$ The radius is: $r^{2}=25$ $\sqrt{r^{2}}=\sqrt{25}$ $r=5$ The graph is:
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