## Precalculus (6th Edition) Blitzer

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.