Let $A=(0,4)$  and  $B=(2\cos \theta ,2\sin \theta )$, for some  $0<\theta <\frac{\pi }{2}$. Let  $P$ divide the line segment  $AB$  in the ratio  $2:3$ internally. The locus of  $P$ is

Let P be  $(x,y)$. Then
 $(x,y)=(\frac{4\cos \theta +0}{5},\frac{4\sin \theta +12}{5})$
$\therefore x=\frac{4\cos \theta }{5}$    and $y=\frac{4\sin \theta +12}{5}$ 
$\Rightarrow x=\frac{4}{5}\cos \theta$   and  $y-\frac{12}{5}=\frac{4}{5}\sin \theta$
 $$\begin{gathered} \therefore x^2+{(y-\frac{12}{5})}^2=\frac{16}{5}{\cos }^2\theta +\frac{16}{25}{\sin }^2\theta \\ \Rightarrow x^2+{(y-\frac{12}{5})}^2=\frac{16}{25} \end{gathered}$$
Which is equation of a circle.