Question

A rod PQ of length 2a slides with its ends on the axes then locus of circumference of DOPQ is.

• A. $x^2+y^2={2a}^2$
• B. $x^2+y^2={4a}^2$
• C. $x^2+y^2={3a}^2$
• D. $x^2+y^2=a^2$

Answer 1

The correct option is D $x^2+y^2=a^2$

Given length of rod$=2a$

It slides on axis.

Let at a instant the coordinates of end of rods are $(x,0)$ and $(0,y)$

It is making a right angle triangle with origin

Here length of hypotenuse $=2a$

By Pythagoras theorem,

$x^2+y^2=(2a{)}^2$

$x^2+y^2=4a^2$ ……….$\left(1\right)$

As it is a right angle triangle circumcircle of $\Delta OXY$ is the mid point of hypotenuse $(\frac{x}{2},\frac{y}{2})$

$\therefore x=\frac{x}{2},y=\frac{y}{2}$

$x=2x$ $y=2y$

Substitute in $\left(1\right)$

$(2x{)}^2+(2y{)}^2=4a^2$

$4[x^2+y^2]=4a^2$

$x^2+y^2=a$.