Question

Show that the relation $R$ in the set $A$ of points in a plane given by $R=\left\{\right(P,Q):\text{distance of the point}P\text{from the origin is same as the distance of the point}Q\text{from the origin}\}$, 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 origin as centre.

Answer 1

$R=\left\{\right(P,Q):\text{distance of point P from the origin is the same as the distance of point Q from the origin}\}$
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 the same as the distance of point $Q$ from the origin.
$\Rightarrow$ The distance of point $Q$ from the origin is the same as the distance of point $P$ from the origin.
$\Rightarrow (Q,P)\in R$
$\therefore R$ is symmetric.
Now, let $(P,Q),(Q,S)\in R$.
$\Rightarrow$ 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.
$\Rightarrow$ The distance of points $P$ and $S$ from the origin is the 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 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 $P$.