## Calculus: Early Transcendentals 8th Edition

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.
$\{(x,y)\vert~x^2+y^2 \leq 1 ~\}$ 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.