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 $P\ne (0,0)$ is the circle passing through P with the origin as centre.

Answer 1

Here, R ={(P, Q)}: distance of point P from the origin is same as the distance of point Q from the origin.

Clearly, $(P,P)\in R$, since the distance of point P from the origin is always the same as the distance of the same point P from the origin. Therefore, R is reflexive.
Now, let $(P,Q)\in R$
$\Rightarrow$ The distance of point P from the origin is same as the distance of point Q from the origin.
$\Rightarrow$ The distance of point Q from the origin is same as the distance of point P from the origin.
$\Rightarrow (Q,P)\in R$
Therefore, R is symmetric.
Now, let $(P,Q),(Q,S)\in R$
The distance of point P and Q from the origin is same and also the distance of points Q and S from the origin is same.
$\Rightarrow$ The distance of points P and S from the origin is same.
$\Rightarrow (P,S)\in R$
Therefore, R is transitive. Therefore, R is an equivalence relation.
The set of all points related to $P\ne (0,0)$ 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 this circle passes through point P.