Question

graph the inequality.
$x^{2}+y^{2}>4$

Explanation:

Step1: Recall the equation of a circle

The standard equation of a circle is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. For \(x^{2}+y^{2}=4\), the center is \((0,0)\) and \(r = 2\) (since \(r^{2}=4\)).

Step2: Determine the boundary line

Since the inequality is \(x^{2}+y^{2}>4\), the boundary line \(x^{2}+y^{2}=4\) is a dashed circle (because the points on the circle do not satisfy the strict inequality \(>\)).

Step3: Test a point

Test the point \((0,0)\): Substitute \(x = 0\) and \(y = 0\) into \(x^{2}+y^{2}>4\). We get \(0^{2}+0^{2}=0
ot>4\). So the region that does not contain the origin \((0,0)\) is the solution set.

Answer:

The graph should be a dashed - circle centered at the origin \((0,0)\) with radius \(r = 2\), and the region outside the circle should be shaded.