A rod of length $6$ units slides with its ends on the coordinate axes. The locus of the midpoint of the rod is

Let $A\left(a,0\right) \& B\left(0,b\right)$ are the end points of the rod and $P\left(h,k\right)$ be the mid-point.

So, $\frac{a}{2}=h \& \frac{b}{2}=k$

$\Rightarrow a=2h \& b=2k$

Length of the rod is given $6$ units

$\Rightarrow \sqrt{{\left(0-a\right)}^2+{\left(b-0\right)}^2}=6$

$\Rightarrow a^2+b^2=6^2$

$\Rightarrow {\left(2h\right)}^2+{\left(2k\right)}^2=6^2$

$\Rightarrow 4h^2+4k^2=36$

So, locus of $P$ is $x^2+y^2=9$