Answer
The graph is shown below:
Work Step by Step
Put the equal symbol in place of the inequality symbol:
Now, the equation is ${{\left( x-2 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=9$.
For a circle having center $\left( h,k \right)$ and radius $ r $, the general equation of the circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
So, the equation ${{\left( x-2 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=9$ can be represented as ${{\left( x-2 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{3}^{2}}$
Thus, the center of the circle $\left( h,k \right)=\left( 2,4 \right)$ and radius $ r=3$.
To plot this circle in the rectangular coordinate system, consider $\left( 2,4 \right)$ as the origin and draw a circle taking 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.
Now, consider a test point $\left( 0,0 \right)$ and check whether the test point satisfies the inequality
${{\left( x-2 \right)}^{2}}+{{\left( y-4 \right)}^{2}}>9$
$\begin{align}
{{\left( 0-2 \right)}^{2}}+{{\left( 0-4 \right)}^{2}}\overset{?}{\mathop{>}}\,9 & \\
4+16\overset{?}{\mathop{>}}\,9 & \\
20>9 & \\
\end{align}$
It is correct. Therefore, the test point satisfies the inequality, so we shade the plane containing the test point and draw the circle with a dotted line as it contains only the greater-than sign.
Thus, the graph of this inequality is the shaded region.
