## Precalculus (6th Edition) Blitzer

Let us consider the inequality ${{x}^{2}}+{{y}^{2}}>4$; substitute the equals symbol in place of the inequality and rewrite the equation as given below: ${{x}^{2}}+{{y}^{2}}=4$ To find the value of the x-intercept, substitute y = 0 as given below: \begin{align} & {{x}^{2}}+{{y}^{2}}=4 \\ & {{x}^{2}}+{{0}^{2}}=4 \\ & x=\pm 2 \end{align} To find the value of the y-intercept, substitute x = 0 as given below: \begin{align} & {{x}^{2}}+{{y}^{2}}=4 \\ & {{0}^{2}}+{{y}^{2}}=4 \\ & y=\pm 2 \end{align} The intercepts represent that it is a circle with origin $\left( 0,0 \right)$ and radius $2$. 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}^{2}}+{{0}^{2}}>4 \\ & 0>4 \end{align} So, the above expression is incorrect, which means the shaded region will not contain the test point; that is, the region away from the origin will be shaded. And the circle passes through four points $\left( 2,0 \right),\left( 0,2 \right),\left( -2,0 \right)$ and $\left( 0,-2 \right)$. Plot the graph using the intercepts as given below: Thus, the graph of the provided inequality is plotted and the circle passes through the points $\left( 2,0 \right),\left( 0,2 \right),\left( -2,0 \right)$, and $\left( 0,-2 \right)$. Also, we see that the shaded region does not contain the origin.