Show that the relation R in the set A of points in a plane given by R $= \{(P, Q) :$ distance of the point P rom 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.
R $= \{(P, Q)$: distance of the point P from the origin is same as the distance of the point Q from the origin$\}$
Let $P({x_1},\;{y_1}),\;Q({x_2},\;{y_2})\;\;and\;\;O(0,\;0).$
Therefore, $OP = OQ = \sqrt {x_1^2 + y_1^2} = \sqrt {x_2^2 + y_2^2} \Rightarrow x_1^2 + y_1^2 = x_2^2 + y_2^2$
(i) Reflexive
$P \in A \Rightarrow (P,\;P) \in R$ $(\therefore \;\;\;OP = OP)$
Distance of the point P from origin is same as the distance of the point P from origin.
Therefore, R is reflexive.
(ii) Symmetric
$P,\;Q \in A;\;\;(P,\;Q) \in R$
$\Rightarrow$ Distance of the point P from origin is same as the distance of point Q from origin
i.e., OP $=$ OQ
$\Rightarrow$ OQ $=$ OP $\Rightarrow$ $(Q,\;P) \in R$ Therefore, R is symmetric.
(iii) Transitive
$P,\;Q,\;S \in R,\;\;(P,\;Q) \in R\;\;\;and\;\;(Q,\;S) \in R,$
$\Rightarrow$ $OP = OQ$ and $OQ = OS \Rightarrow OP = OS \Rightarrow (P,\;S) \in R$
$\Rightarrow$ R is transitive. Hence, R is an equivalence relation.
We have to find the set of points related to $P \ne (0,\;0)$
As, $x_1^2 + y_1^2 = x_2^2 + y_2^2 = {r^2} \Rightarrow {x^2} + {y^2} = {r^2}$
Which represents a circle with centre (0, 0) and radius $=$ r.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Relations and Functions. Curated by Sachin Sharma. Free for all students.