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: 61

Answer

See below:
1569143346

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