Precalculus (6th Edition) Blitzer

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.