A rod of length 2l slides with its ends on two perpendicular lines, then the locus of its mid-point is

• Correct answer
  A

  $x^2+y^2=l^2$
• B

  $x^2-y^2=l^2$
• C

  $2x^2+2y^2=l^2$
• D

  $2x^2-2y^2=l^2$

Solution

Based on given information, diagram is drawn. Let D be the point on rod. Let $x=2 \alpha$ and $y=2 \beta$
In $\triangle A O B$, $$ \begin{aligned} x^2+y^2 & =(2 l)^2 \\ (2 \alpha)^2+(2 \beta)^2 & =4 l^2 \\ \alpha^2+\beta^2 & =l^2 \\ \frac{x^2}{4}+\frac{y^2}{4} & =l^2 \end{aligned} $$ $x^2+y^2=4 l^2 \quad\left\{\begin{array}{l}\because x=2 \alpha \Rightarrow \alpha=x / 2 \\ y=2 \beta \Rightarrow \beta=y / 2\end{array}\right.$

Hence, the correct answer is option: 'A'.