A stick of length r units slides with its ends on coordinate axes. Then the locus of the midpoint of the…

Let the stick intercepts $x$-axis at point $\left(a,0\right)$ and $y$-axis at point $\left(0,b\right)$ and $\left(\alpha ,\beta \right)$ are midpoints of the stick. then,

$\frac{a+0}{2}=\alpha \Rightarrow a=2\alpha$

$\frac{b+0}{2}=\beta \Rightarrow b=2\beta$

Now given length of stick is $r$ units,

Then, $r^2={\left(a-0\right)}^2+{\left(0-b\right)}^2$

$\Rightarrow r^2=a^2+b^2$

$\Rightarrow r^2={\left(2\alpha \right)}^2+{\left(2\beta \right)}^2$

$\Rightarrow r^2=4{\alpha }^2+4{\beta }^2$

${\alpha }^2+{\beta }^2={\left(\frac{r}{2}\right)}^2$

If we replace $\left(\alpha ,\beta \right)$ by $\left(x,y\right)$ then, our equation becomes,

$x^2+y^2={\left(\frac{r}{2}\right)}^2$

So, locus will be a circle with center at origin and radius $\frac{r}{2}$,

Hence length is given by $2\mathrm{\pi }\times \frac{\mathrm{r}}{2}=\mathrm{\pi r}$.