The circle is defined as the locus of a point equidistant from a fixed point.
If we say the center is at $(a,b)$ then the locus of a point $(x,y)$ at a distance $r$ creates a curve (a boundary of length $2\pi r$) which bounds an inner area of $\pi r^2$
Then, this circle, on the Cartesian plane can be represented by the equation:
$$(x-a)^2 + (y-b)^2 = r^2$$
Figure $1$ shows a circle. It can be noted that point $Z$ satisfies the given equation while point $P$ does not satisfy. Hence, we say $Z$ lies on the circle.
Note: It is not wrong to say that $P$ lies in the circle.
Now, what is Figure $2$? It is a disk (also spelled disc).
A disk is defined as the region in a plane bounded by a circle.
Both $X$ and $Q$ clearly lie on the disk.
But there is a problem when thinking about figure $2$.
It is clearly intended that $X$ lies on the periphery of the darkened area.
But does this mean that $X$ lies on the circle corresponding to the disk?
I express this as a problem because the area inside the circle defined earlier can be denoted as:
$$(x-a)^2 + (y-b)^2 \lt r^2$$
Now, point $Q$ satisfies this but it is unclear whether $X$ satisfies it.
But don't be confused, remember the remark I made earlier. "Both $X$ and $Q$ clearly lie on the disk."
This is because a disk is conventionally taken to be a closed disk (unless mentioned otherwise to be open). That is, the disk is defined as the set of all points which satisfy:
$$ (x-a)^2 + (y-b)^2 \le r^2$$
Hopefully, my answer was helpful :D