#### Answer

The graph is shown below:

#### Work Step by Step

Let us consider the inequalities and put the equals symbol in place of the inequality and rewrite the equation as given below:
${{x}^{2}}+{{y}^{2}}=1$
${{x}^{2}}+{{y}^{2}}=4$
The first equation ${{x}^{2}}+{{y}^{2}}=1$ represents a circle with origin at $\left( 0,0 \right)$ and radius 1 that can be rewritten as ${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{1}^{2}}$.
Plot the graph using the equation of this circle in the rectangular coordinate system with origin at $\left( 0,0 \right)$ and radius $ r=1$.
Then, this circle divides the plane into three regions: the circle itself, the inner plane of the circle, and the outer plane of the circle.
Now, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}>1 \\
& 0+0\overset{?}{\mathop{>}}\,1 \\
& 0<1 \\
\end{align}$
Since the test point does not satisfy the inequality, shade the outer plane of the circle, that is, away from the origin $\left( 0,0 \right)$.
The second equation ${{x}^{2}}+{{y}^{2}}=4$ represents a circle with origin at $\left( 0,0 \right)$ and radius 2 that can be rewritten as ${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{2}^{2}}$.
Plot the graph using the equation of this circle in the rectangular coordinate system with origin at $\left( 0,0 \right)$ and radius $ r=2$.
Now, this circle divides the plane into three regions: the circle itself, the inner plane of the circle, and the outer plane of the circle.
Then, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}<4 \\
& 0+0<4 \\
& 0<4
\end{align}$
Since the test point satisfies the inequality, shade the inner plane of the circle, that is, towards the origin $\left( 0,0 \right)$.
Thus, plot the graph of the inequalities using the common shaded region.