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

(N/A) $R = \{(P, Q) : \text{Distance of point } P \text{ from the origin is the same as the distance of point } Q \text{ from the origin}\}$.
$1$. Reflexivity: For any point $P \in A$,the distance of $P$ from the origin is equal to the distance of $P$ from the origin. Thus,$(P, P) \in R$. Therefore,$R$ is reflexive.
$2$. Symmetry: Let $(P, Q) \in R$. This means the distance of $P$ from the origin is the same as the distance of $Q$ from the origin. This implies the distance of $Q$ from the origin is the same as the distance of $P$ from the origin. Thus,$(Q, P) \in R$. Therefore,$R$ is symmetric.
$3$. Transitivity: Let $(P, Q) \in R$ and $(Q, S) \in R$. This means the distance of $P$ from the origin equals the distance of $Q$ from the origin,and the distance of $Q$ from the origin equals the distance of $S$ from the origin. Consequently,the distance of $P$ from the origin equals the distance of $S$ from the origin. Thus,$(P, S) \in R$. Therefore,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
For the second part,the set of all points related to $P \neq (0, 0)$ consists of all points $Q$ such that the distance of $Q$ from the origin is equal to the distance of $P$ from the origin. Let $OP = k$. Then all such points $Q$ lie at a constant distance $k$ from the origin. This is the definition of a circle with the origin as the center and radius $k$,which passes through $P$.