Answer
Describe the set of all points that correspond to the given equations.
Work Step by Step
\[
\begin{aligned}
\left\|-\vec{r}_{0}+\vec{r}\right\|=1 & \Rightarrow\left\|\left\langle -x_{0}+x, -y_{0}+y\right\rangle\right\|=1 \\
& \Rightarrow \sqrt{\left(-x_{0}+x\right)^{2}+\left(-y_{0}+y\right)^{2}}=1 \\
& \Rightarrow\left(-x_{0}+x\right)^{2}+\left(-y_{0}+y\right)^{2}=1
\end{aligned}
\]
This equation corresponds to the set of all points in a space of 2 over a radius of $1=r$ centered at $C\left(x_{0}, y_{0}\right)$
$b)$
Since $\left\|\vec{r}-\vec{r}_{0}\right\| \geqslant 0$ for all vectors $\vec{r}_{0}$ and $\vec{r}$, we get that
\[
\left\|-\vec{r}_{0}+\vec{r}\right\| \leqslant 1 \Rightarrow\left(-x_{0}+x\right)^{2}+\left(-y_{0}+y\right)^{2} \leq 1
\]
Thus, the inequality corresponds to the set of all points in 2d-space that are at a distance $d \leq 1$ from $C\left(x_{0}, y_{0}\right) .$ In other words, a set of all points inside a circle.
c) The inequality
\[
\left\|-\vec{r}_{0}+\vec{r}\right\|>1 \Rightarrow\left(-x_{0}+x\right)^{2}+\left(-y_{0}+y\right)^{2}>1
\]
This corresponds to the set of all points outside the circle.
