Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 280: 62

Answer

See below:
1569143374

Work Step by Step

We know that for a circle the standard form is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$, Where the center is $\left( h,k \right)$ and the radius is $r$. So, the equation can be written as: $\begin{align} & {{x}^{2}}+{{y}^{2}}+x+y-\frac{1}{2}=0=0 \\ & \left( {{x}^{2}}+x \right)+\left( {{y}^{2}}+y \right)=\frac{1}{2} \\ & \left( {{x}^{2}}+x+\frac{1}{4} \right)+\left( {{y}^{2}}+y+\frac{1}{4} \right)=\frac{1}{2}+\frac{1}{4}+\frac{1}{4} \\ & {{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( y+\frac{1}{2} \right)}^{2}}=1 \end{align}$ So, the equation of the circle in the standard form is given as: ${{\left( x-\left( -\frac{1}{2} \right) \right)}^{2}}+{{\left( y-\left( -\frac{1}{2} \right) \right)}^{2}}={{1}^{2}}$ Now compare this equation to the standard form, to get the value of $h=-\frac{1}{2},k=-\frac{1}{2},\,\,\text{ and }\,\,r=1$. Therefore, the center is $\left( -\frac{1}{2},-\frac{1}{2} \right)$ and the radius is $1$ unit.
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