Consider O as the origin of the plane

So R = {(P, Q): OP = OQ)

By considering properties of relation R

Symmetric:

Consider P and Q as the two points in set A where (P, Q) ∈ R

We can write it as

OP = OQ where (Q, P) ∈ R

So we get (P, Q) ∈ R and (Q, P) ∈ R for P, Q ∈ A

Hence, R is symmetric.

Reflexivity:

Consider P as any point in set A where OP = OP

We know that (P, P) ∈ R for all P ∈ A

Hence, R is reflexive.

Transitivity:

Consider P, Q and S as three points in a set A where (P, Q) ∈ R and (Q, S) ∈ R

We know that OP = OQ and OQ = OS

So we get OP = OS where (P, S) ∈ R

Hence, R is transitive

Therefore, R is an equivalence relation.

Consider P as a fixed point in set A and let Q be a point in set A where (P, Q) ∈ R

We know that OP = OQ where Q moves in the plane that its distance from O.

So we get O (0, 0) = OP

So the locus of Q is a circle having centre at O and OP as the radius

Therefore, the set of all points which is related to P passes through the point P having O as centre.