Chapter 1.8, Problem 110E

To determine

To graph: The region provided by the set $\left\{\left(x,y\right)|x^2+y^2>4\right\}$ .

Explanation of Solution

Given information:

The set, $\left\{\left(x,y\right)|x^2+y^2>4\right\}$ .

Graph:

The graph of the region provided by the set $\left\{\left(x,y\right)|x^2+y^2>4\right\}$ is sketched in the coordinate plane.

Consider the equation $x^2+y^2=4$ .

Rewrite the equation as,

  ${\left(x-0\right)}^2+{\left(y-0\right)}^2=2^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-0\right)}^2+{\left(y-0\right)}^2=2^2$ .

Here, $h=0,k=0\text{ and }r=2$ .

Therefore, center of circle is $\left(0,0\right)$ and radius is $2$ .

Now, on the Cartesian plane construct a circle with center at origin and radius 2 units.

Since, the inequality is $\left\{\left(x,y\right)|x^2+y^2>4\right\}$

Now, shade the common region that lie otiside the circle..

The region obtained is provided below,

  

Interpretation:

The region provided by the set $\left\{\left(x,y\right)|x^2+y^2>4\right\}$ is shaded region. The region shaded forms a circle with center at origin and radius 2 units. The broken line on boundary of circle denote that the boundary is not a part of the shaded region. The region described by inequality is region that lie outside the circle and all such points in coordinate plane that satisfy $x^2+y^2>4$ .