Question

If 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, then the set of all points related to a point $P \neq(0,0)$ is the

• A. circle
• B. ellipse
• C. parabola
• D. hyperbola

Answer 1

Answer: (A)

Here, $R=\{(P, Q)$ : distance of point $P$ from the origin is same as the distance of point $Q$ from the origin . Clearly, $(P, P) \in R$, since the distance of point $P$ from the origin is always the same as the distance of the same point $P$ from the origin.
Therefore, $R$ is reflexive. Now, let $(P, Q) \in R$.
$\Rightarrow$ The distance of point $P$ from the origin is same as the distance of point $Q$ from the origin.
$\Rightarrow$ The distance of point $Q$ from the origin is same as the distance of point $P$ from the origin.
$\Rightarrow(Q, P) \in R$
Therefore, $R$ is symmetric.
Now, let $(P, Q),(Q, S) \in R$
The distance of points $P$ and $Q$ from the origin is same and also the distance of points $Q$ and $S$ from the origin is same.
$\Rightarrow$ The distance of points $P$ and $S$ from the origin is same.
$\Rightarrow(P, S) \in R$
Therefore, $R$ is transitive. Hence, $R$ is an equivalence relation (which is already given). The set of all points related to $P \neq(0,0)$ will be those points whose distance from the origin is the same as the distance of point $P$ from the origin.
In other words, if $O(0,0)$ is the origin and $O P=k$, then the set of all points related to $P$ is the set of those points which are at a distance of $k$ from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point $P$.