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: 63

Answer

See below:
1569143401

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: $\begin{align} & {{x}^{2}}+{{y}^{2}}+3x-2y-1=0 \\ & \left( {{x}^{2}}+3x \right)+\left( {{y}^{2}}-2y \right)=1 \\ & \left( {{x}^{2}}+3x+\frac{9}{4} \right)+\left( {{y}^{2}}-2y+1 \right)=1+\frac{9}{4}+1 \\ & {{\left( x+\frac{3}{2} \right)}^{2}}+{{\left( y-1 \right)}^{2}}=\frac{17}{4} \end{align}$ So, the equation of the circle in the standard form is given as: ${{\left( x-\left( -\frac{3}{2} \right) \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( \frac{\sqrt{17}}{2} \right)}^{2}}$ Now, compare this equation to the standard form, to get the value of $h=-\frac{3}{2},k=1,\,\,\text{ and }\,\,r=\frac{\sqrt{17}}{2}$. Therefore, the center is $\left( -\frac{3}{2},1 \right)$ and the radius is $\frac{\sqrt{17}}{2}$ units.
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