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.