Show that the relation `R` on 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!=(0,\ 0)` is the circle passing through `P` with origin as centre.

Explanation

`R= {(P, Q): ` distance of point P from origin is same as the distance of point Q from origin }
Now (P, P) `in R` is always true.
`therefore R` is reflexive.
Let (P, Q)` in `R
`rArr ` distance of P from origin = distance of Q from origin.
`rArr ` distance of Q from origin = distance of P from origin.
`rArr (Q, P) in R`
`therefore R ` is symmetric.
Let (P, Q) `in R and (Q, S) in R`
`rArr ` distance of P from origin = distance of Q from origin.
and distance of Q from origin = distance of S from origin
`rArr ` distance of P from origin = distance of S from origin.
`rArr " " (P, S) in R`
`therefore R `is transitive.
`because R` is reflexive, symmetric and transitive.
`therefore` R is an equivalence relation.
Now take that point related to point `P ne (0, 0)` whose distance from origin is equal to distance of point P from origin. So if O(0, 0) is origin and OP =k then the points related to point `P ne (0, 0)` will be at k distance from origin. So the set of points related to point P is a circle whose centre is origin. This circles passes through the point P.