Answer
See below:
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.
