Question

Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all point related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Answer 1

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) ∈ 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) ∈ R
⇒ The distance of point P from the origin is the same as the distance of point Q from the
origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the
origin.
⇒ (Q, P) ∈ R
∴R is symmetric.

Now,
Let (P, Q), (Q, S) ∈ R.
⇒ 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.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ 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.