$\mathrm{R} =\{( \mathrm{P} , \mathrm{Q} ):$ Distance of point $\mathrm{P} $ from the origin is the same as the distance of point $\mathrm{Q}$ from the origin $\}$

Clearly, $(\mathrm{P}. \mathrm{P}) \in \mathrm{R}$ since the distance of point $\mathrm{P}$ from the origin is always the same as the distance of the same point $\mathrm{P}$ from the origin.

$\therefore \mathrm{R}$ is reflexive.

Now, Let $(\mathrm{P},\, \mathrm{Q}) \,\in \mathrm{R}$

$\Rightarrow$ The distance of point $\mathrm{P}$ from the origin is the same as the distance of point $\mathrm{Q}$ from the origin.

$\Rightarrow $ The distance of point $\mathrm{Q}$ from the origin is the same as the distance of point $\mathrm{P}$ from the origin.

$\Rightarrow$  $(\mathrm{Q}, \mathrm{P}) \in \mathrm{R}$

$\therefore \mathrm{R}$ is symmetric.

Now, Let $( \mathrm{P} ,\, \mathrm{Q} ),\,( \mathrm{Q} , \,\mathrm{S} ) \in \mathrm{R}$

$\Rightarrow$ The distance of points $\mathrm{P}$ and $\mathrm{Q}$ from the origin is the same and also, the distance of points $\mathrm{Q}$ and $\mathrm{S}$ from the origin is the same.

$\Rightarrow$ The distance of points $\mathrm{P}$ and $\mathrm{S}$ from the origin is the same.

$\Rightarrow$    $( \mathrm{P} , \,\mathrm{S} ) \in \mathrm{R}$

$\therefore \mathrm,{R}$ is transitive.

Therefore, $\mathrm{R}$ is an equivalence relation.

The set of all points related to $\mathrm{P} \neq(0,0)$ will be those points whose distance from the origin is the same as the distance of point $\mathrm{P}$ from the origin.

In other words, if $\mathrm{O}(0,0) $ is the origin and $\mathrm{OP} = \mathrm{k}$, then the set of all points related to $\mathrm{P}$ is at a distance of $\mathrm{k}$ from the origin.

Hence, this set of points forms a circle with the centre as the origin and this circle passes through point $\mathrm{P}$.