#### Answer

The set includes all the points inside the circle $x^2+y^2 = 1$
Note that the points on the circle are also included in the set.

#### Work Step by Step

$\{(x,y)\vert~x^2+y^2 \leq 1 ~\}$
We can write the general equation for a circle:
$(x-a)^2+(y-b)^2 = r^2$
where $(a,b)$ is the center of the circle and $r$ is the radius
The set includes all the points inside the circle $x^2+y^2 = 1$
The center of the circle is $(0,0)$ and the radius of the circle is $1$
Note that the points on the circle are also included in the set.