Question

Let $R=\left\{\left(P,Q\right)\right|P,Q$ are at the same distance from the origin} , then the equivalence class of $\left(1,-1\right)$ is the set :

• A. $S=\left\{\left(x,y\right)|x^2+y^2=1\right\}$
• B. $S=\left\{\left(x,y\right)|x^2+y^2=4\right\}$
• C. $S=\left\{\left(x,y\right)|x^2+y^2=\sqrt{2}\right\}$
• D. $S=\left\{\left(x,y\right)|x^2+y^2=2\right\}$

Answer 1

The correct option is D

$S=\left\{\left(x,y\right)|x^2+y^2=2\right\}$

Explanation for correct answer:

Let us assume the co-ordinates of point $P$ be $\left(c,d\right)$ while the co-ordinates of point $Q$ be $(l,m)$.

Given, $P$ and $Q$ are at same distance from the origin, i.e $OP\hspace{0.17em}=OQ\dots \left(1\right)$

Origin, $O=(0,0)$

Using distance formula, $d=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

$$\begin{gathered} \Rightarrow \sqrt{{\left(c-0\right)}^2+{\left(d-0\right)}^2}=\sqrt{{\left(l-0\right)}^2+{\left(m-0\right)}^2} \\ \Rightarrow \sqrt{c^2+d^2}=\sqrt{l^2+m^2} \end{gathered}$$

Squaring both side, we get,

$c^2+d^2=l^2+m^2\dots \left(2\right)$

Let $R=(x,y),S=\left(1,-1\right)$

$OR\hspace{0.17em}=OS$ (equivalence class)

Put the value of $R$ and $S$ in equation $2$, we get,

$x^2+y^2=2$

Thus, the set is $x^2+y^2=2$.

Hence, the correct option is (D)