Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises - Page 135: 104

Answer

$a)$ Yes, the equation represents a circle $b)$ Center: $\Big(\dfrac{1}{2},-2\Big);$ radius: $\dfrac{3\sqrt{2}}{2}$

Work Step by Step

$2x^{2}+2y^{2}-2x+8y=\dfrac{1}{2}$ $a)$ Take out common factor $2$ from the left side: $2(x^{2}+y^{2}-x+4y)=\dfrac{1}{2}$ Take the $2$ to divide the right side: $x^{2}+y^{2}-x+4y=\dfrac{1}{(2)(2)}$ $x^{2}+y^{2}-x+4y=\dfrac{1}{4}$ On the left side, group the terms with $x$ together and the terms with $y$ together: $(x^{2}-x)+(y^{2}+4y)=\dfrac{1}{4}$ Complete the square for each group. To do so, add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and do it for both groups. For the first group, $b=-1$ and for the second group, $b=4$ $\Big[x^{2}-x+\Big(\dfrac{-1}{2}\Big)^{2}\Big]+\Big[y^{2}+4y+\Big(\dfrac{4}{2}\Big)^{2}\Big]=\dfrac{1}{4}+\Big(\dfrac{-1}{2}\Big)^{2}+\Big(\dfrac{4}{2}\Big)^{2}$ $\Big(x^{2}-x+\dfrac{1}{4}\Big)+\Big(y^{2}+4y+4\Big)=\dfrac{1}{4}+\dfrac{1}{4}+4$ $\Big(x^{2}-x+\dfrac{1}{4}\Big)+\Big(y^{2}+4y+4\Big)=\dfrac{9}{2}$ Now, each group is a perfect square trinomial. Factor them both: $\Big(x-\dfrac{1}{2}\Big)^{2}+\Big(y+2\Big)^{2}=\dfrac{9}{2}$ Since it was possible to represent the original equation in the standard form of the equation of a circle, then it represents a circle. $b)$ The equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius. Obtain the center and the radius from the equation found in part $a$ of this problem: $\Big(x-\dfrac{1}{2}\Big)^{2}+\Big(y+2\Big)^{2}=\dfrac{9}{2}$ radius: $r=\sqrt{\dfrac{9}{2}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$ Center: $\Big(\dfrac{1}{2},-2\Big);$ radius: $\dfrac{3\sqrt{2}}{2}$
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