#### Answer

The graph is shown below:

#### Work Step by Step

Since the equation is of a circle, make the equation into the general form of a circle $\left( x-{{h}^{2}} \right)+{{\left( y-k \right)}^{2}}={{r}^{2}}$. The center of the circle is expressed by the coordinates $\left( h,k \right)$ and r represents the radius of the circle.
Simplify:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}-2x+4y=4 \\
& {{x}^{2}}-2x+1+{{y}^{2}}+4y=5 \\
& {{\left( x-1 \right)}^{2}}+{{y}^{2}}+4y=5
\end{align}$
Also,
$\begin{align}
& {{\left( x-1 \right)}^{2}}+{{y}^{2}}+4y+4=5+4 \\
& {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9
\end{align}$
Thus, the center is $\left( 1,-2 \right)$ and the radius is 3.
Equation ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9$ can be expressed as $y=\sqrt{9-{{\left( x-1 \right)}^{2}}}-2$.
Substitute $x=1$ in the above equation:
$\begin{align}
& y=\sqrt{9-{{\left( 1-1 \right)}^{2}}}-2 \\
& =\pm 3-2 \\
& =1,-5
\end{align}$
At $x=4$ ,
$\begin{align}
& y=\sqrt{9-{{\left( 4-1 \right)}^{2}}}-2 \\
& =\sqrt{9-9}-2 \\
& =-2
\end{align}$
At $x=-2$ ,
$\begin{align}
& y=\sqrt{9-{{\left( -2-1 \right)}^{2}}}-2 \\
& =\sqrt{9-{{\left( -3 \right)}^{2}}}-2 \\
& =\sqrt{9-9}-2 \\
& =-2
\end{align}$
Thus, the points $\left( 1,1 \right),\left( 1,-5 \right),\left( 4,-2 \right)$ and $\left( -2,-2 \right)$ satisfy the $y=\sqrt{9-{{\left( x-1 \right)}^{2}}}-2$.