Q.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 points related to a point $P \ne (0, 0)$ is the circle passing through P with origin as centre.

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

The three checks

1. Reflexive — show $(a, a) \in R$ for every $a \in A$.
2. Symmetric — assume $(a, b) \in R$ and deduce $(b, a) \in R$.
3. Transitive — assume $(a, b) \in R$ and $(b, c) \in R$, and deduce $(a, c) \in R$.

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$.

Reflexive: $a - a = 0$, and $0$ is divisible by $5$, so $a\,R\,a$ for every integer $a$. ✓

Symmetric: if $a\,R\,b$, then $a - b = 5k$ for some integer $k$. Then $b - a = -5k = 5(-k)$, also a multiple of $5$, so $b\,R\,a$. ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$, then $a - b = 5k$ and $b - c = 5m$. Adding, $a - c = 5(k + m)$, a multiple of $5$, so $a\,R\,c$. ✓

All three hold, so $R$ is an equivalence relation. …