A simple pendulum of length $\mathcal{l}$  is suspended by a nail on a vertical wall at point O. Another nail is to be fixed on the wall such that if the pendulum is released from its initial horizontal position, it just completes a vertical circle round that nail as shown. The locus of the point where the second nail can be fixed is $k(x^2+y^2)={(3\mathcal{l}-2y)}^2$. Find the value of  $k$.

Let the second nail be located as a distance r from the first nail and angle $\theta$ below horizontal as shown. Now speed of the bob at its lowest position can be found using conservation of energy.
$\frac{1}{2}m{v}^2=mg[r\sin \theta +\mathcal{l}-r]$
$\Rightarrow v^2=2g(r\sin \theta +\mathcal{l}-r)$ …….(i)
The particle will complete circle about the second nail if   $v^2=5g(\mathcal{l}-r)$……(ii)
From (i) and (ii)
$2(r\sin \theta +\mathcal{l}-r)=5(\mathcal{l}-r)$ 

 $\Rightarrow 3r+2r\sin \theta =3\mathcal{l}$ …….(iii)
Now coordinates of the nail are
$x=r\cos \theta ,y=r\sin \theta$
Using these in (iii), we get
$3\sqrt{x^2+y^2}+2y=3\mathcal{l}$
Or  $9(x^2+y^2)={(3\mathcal{l}-2y)}^2$