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 w the center at the origin.