#### Answer

The graph is shown below:

#### Work Step by Step

Let us consider the inequalities; put the equals symbol in place of the inequality and rewrite the equation as given below:
$ y=1-{{x}^{2}}$
${{x}^{2}}+{{y}^{2}}=9$.
To plot the graph of $ y=1-{{x}^{2}}$, evaluate the value of the y variable for each value of the x variable.
To find the value of the y-intercept, substitute $x = 0$ as given below:
$\begin{align}
& y=1-{{\left( 0 \right)}^{2}} \\
& =1-0 \\
& =1
\end{align}$
To find the value of the x-intercept, substitute $y = 0$ as given below:
$\begin{align}
& 0=1-{{x}^{2}} \\
& {{x}^{2}}=1 \\
& x=\pm 1
\end{align}$
Thus, plot the intercepts $\left( 0,1 \right),\ \left( 1,0 \right)\text{, and }\left( -1,0 \right)$ and join them with a free hand in order to obtain the graph of the equation $ y=1-{{x}^{2}}$.
Then, this parabola divides the plane into three regions: the parabola itself, the inner plane of the parabola, and the outer plane of the parabola.
Now, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade:
$\begin{align}
& y\le 1-{{x}^{2}} \\
& 0\overset{?}{\mathop{\le }}\,1-0 \\
& 0\le 1 \\
\end{align}$
Since the test point satisfies the inequality, shade the inner plane of the parabola, that is, towards the origin $\left( 0,0 \right)$.
For a circle having center $\left( h,k \right)$ and radius $ r $, the general equation of circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
And the second equation ${{x}^{2}}+{{y}^{2}}=9$ represents a circle with origin at $\left( 0,0 \right)$ and radius 3 that can be rewritten as ${{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{3}^{2}}$.
Plot the graph using the equation of this circle in the rectangular coordinate system with origin at $\left( 0,0 \right)$ and radius $ r=3$.
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:
$\begin{align}
& {{x}^{2}}+{{y}^{2}}\le 9 \\
& 0+0\overset{?}{\mathop{\le }}\,9 \\
& 0\le 9 \\
\end{align}$
.
Since the test point satisfies the inequality, shade the inner plane of the circle, that is, towards the origin $\left( 0,0 \right)$.
See the final graph below.