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

For any point $P = Q$, the distance from point $P$ from origin is obviously the same as distance from point $Q$. Hence $R$ is reflexive.

We can also observer that for any two points $P$ and $Q$, if $(P,Q) \in R \Rightarrow (Q,P) \in R$ as the distances from $P$ and $Q$ from origin stay the same and are part of the relation $R$. Hence $R$ is symmetric.

If we take 3 points $P, Q, R$ such that if $P,Q$ are equidistant from the origin, and $Q,R$ are equidistant from the origin, then it follows that $P,R$ must be the same distance also. Hence $R$ is transitive also.

Since $R$ is reflexive, transitive and symmetrical, it is an equivalence relation.

Consider a circle passing through P with centre at $(0,0$) orgin.

Any point $P(p,p)$ where $p \neq 0)$ is equidistant from the origin. We can see that a set of these points all equidistant from the origin form a circle with the centre at the origin.