Answer
We describe each region.
a) $1=x^{2}+y^{2}$
b) $y^{2}+x^{2} \leq 1$
c) $y^{2}+x^{2}>1$
Work Step by Step
Let $\vec {r}=\langle x, y\rangle$ be a random vector within a distance of 2.
a) If $\|\vec{r}\|=1,$ then
\[
\|\vec{r}\|=1 \Rightarrow \sqrt{y^{2}+x^{2}}=1 \Rightarrow 1=y^{2}+x^{2}
\]
This equation corresponds to a radius $1=r$ originally centered.
b) In the same way
\[
\|\vec{r}\| \leq 1 \Rightarrow y^{2}+x^{2} \leq 1
\]
because $\|\vec{r}\| \geq 0$ for all vectors $\vec{r}$. This inequality corresponds to the set of all points within the circle $1=y^{2}+x^{2}$
Finally
\[
\|\vec{r}\|>1 \Rightarrow y^{2}+x^{2} \geq 1
\]
This inequality corresponds to the set of all points in 20 -space that are out of the circle $1=y^{2}+x^{2} .$ In other words, it corresponds to the set of all points located at a distance of $ d> $ 1 from the original.
