Answer
The interior of a circle centered at $(1,0)$, with radius $2$,
including the circle itself.
Work Step by Step
The plane is divided into two regions by the graph of
$(x-1)^{2}+y^{2}=4$
which is a circle centered at $C(1,0)$, with radius $2$.
The two regions are the interior and the exterior of the circle.
Testing ($1,0$) by inserting its coordinates into the inequality,
$0+0\leq 4\qquad $ ... (true)
we see that the center belongs to the solution set.
The curve itself is included, as the inequality $\leq$ is not strict.
The region represented by the inequality:
The interior of a circle centered at $(1,0)$, with radius $2$,
including the circle itself.
