Show that the relation $\mathrm{R}$ in the set $\mathrm{A}$ of points in a plane given by $\mathrm{R} =\{( \mathrm{P} ,\, \mathrm{Q} ):$ distance of the point $\mathrm{P}$ from the origin is same as the distance of the point $\mathrm{Q}$ from the origin $\}$, is an equivalence relation. Further, show that the set of all points related to a point $\mathrm{P} \neq(0,\,0)$ is the circle passing through $\mathrm{P}$ with origin as centre.
Show that the relation $\mathrm{R}$ in the set $\mathrm{A}$ of points in a plane given by $\mathrm{R} =\{( \mathrm{P} ,\, \mathrm{Q} ):$ distance of the point $\mathrm{P}$ from the origin is same as the distance of the point $\mathrm{Q}$ from the origin $\}$, is an equivalence relation. Further, show that the set of all points related to a point $\mathrm{P} \neq(0,\,0)$ is the circle passing through $\mathrm{P}$ with origin as centre.
$\mathrm{R} =\{( \mathrm{P} , \mathrm{Q} ):$ Distance of point $\mathrm{P} $ from the origin is the same as the distance of point $\mathrm{Q}$ from the origin $\}$
Clearly, $(\mathrm{P}. \mathrm{P}) \in \mathrm{R}$ since the distance of point $\mathrm{P}$ from the origin is always the same as the distance of the same point $\mathrm{P}$ from the origin.
$\therefore \mathrm{R}$ is reflexive.
Now, Let $(\mathrm{P},\, \mathrm{Q}) \,\in \mathrm{R}$
$\Rightarrow$ The distance of point $\mathrm{P}$ from the origin is the same as the distance of point $\mathrm{Q}$ from the origin.
$\Rightarrow $ The distance of point $\mathrm{Q}$ from the origin is the same as the distance of point $\mathrm{P}$ from the origin.
$\Rightarrow$ $(\mathrm{Q}, \mathrm{P}) \in \mathrm{R}$
$\therefore \mathrm{R}$ is symmetric.
Now, Let $( \mathrm{P} ,\, \mathrm{Q} ),\,( \mathrm{Q} , \,\mathrm{S} ) \in \mathrm{R}$
$\Rightarrow$ The distance of points $\mathrm{P}$ and $\mathrm{Q}$ from the origin is the same and also, the distance of points $\mathrm{Q}$ and $\mathrm{S}$ from the origin is the same.
$\Rightarrow$ The distance of points $\mathrm{P}$ and $\mathrm{S}$ from the origin is the same.
$\Rightarrow$ $( \mathrm{P} , \,\mathrm{S} ) \in \mathrm{R}$
$\therefore \mathrm,{R}$ is transitive.
Therefore, $\mathrm{R}$ is an equivalence relation.
The set of all points related to $\mathrm{P} \neq(0,0)$ will be those points whose distance from the origin is the same as the distance of point $\mathrm{P}$ from the origin.
In other words, if $\mathrm{O}(0,0) $ is the origin and $\mathrm{OP} = \mathrm{k}$, then the set of all points related to $\mathrm{P}$ is at a distance of $\mathrm{k}$ from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point $\mathrm{P}$.
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