Let $R = {(P, Q) | P$ and $Q$ are at the same distance from the origin} be a relation, then the equivalence class of $(1, −1)$ is the set :
(A)
$S = {(x, y) | x^2 + y^2 = \sqrt{2}}$
(B)
$S = {(x, y) | x^2 + y^2 = 2}$
(C)
$S = {(x, y) | x^2 + y^2 = 1}$
(D)
$S = {(x, y) | x^2 + y^2 = 4}$
MEDIUM
Correct Answer:B
Explanation
Given $R = {(P, Q) | P$ and $Q$ are at the same distance from the origin}.
Then the equivalence class of $(1, -1)$ will contain all such points that lie on the circumference of the circle with the center at the origin and passing through the point $(1, -1)$.
The radius of the circle $= \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$. Therefore, the equation of the circle is $x^2 + y^2 = (\sqrt{2})^2 = 2$.
Required equivalence class of $(S) = {(x, y) | x^2 + y^2 = 2}$.