Show that the relation R in the set A of points in a plane given by

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

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

∴R is reflexive.

Now,

Let $(P, Q) \in R$.

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

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

$\Rightarrow(Q, P) \in R$

∴R is symmetric.

Now,

Let $(P, Q),(Q, S) \in R$.

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

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

$\Rightarrow(P, S) \in R$

∴R is transitive.

Therefore, R is an equivalence relation.

The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.

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

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