$2a = 10 \Rightarrow a = 5 ; 2b = 8 \Rightarrow b = 4$ 
$e^2 = 1 – \frac{{16}}{{25}}\,$=$\frac{9}{{25}}\, \Rightarrow e = \frac{3}{5}\,$ 
Focus = $(3, 0)$ 
Let the circle touches the ellipse at $ P $ and $Q$. Consider a tangent (to both circle and ellipse) at $P$. Let $F$(one focus) be the centre of the circle and other focus be $G.$  A ray from $F$  to $ P$ must retrace its path (normal to the circle). But the reflection propety the ray $FP$  must be reflected along $PG$. This is possible only if $ P, F $ and $G $ are collinear. Thus $P$ must be the end of the major axis.Hence $ r = a – ae = 5 – 3 = 2$ 
  alternately normal to an ellipse at $P$  must pass through the centre $(3, 0)$  of the circle 
$\frac{{ax}}{{\cos \theta }}\,\, – \,\,\frac{{by}}{{\sin \theta }}\,\, = \,\,{a^2} – {b^2} $ $\Rightarrow \frac{{5x}}{{\cos \theta }}\,\, – \,\,\frac{{4y}}{{\sin \theta }}\,\, = \,\,9$ $\left( {\theta \, \ne \,0\,\,or\,\frac{\pi }{2}\,} \right)\,$
$\frac{{15}}{{\cos \theta }}\,\, – \,\,0\, = 9\, \Rightarrow \cos \theta \, = \,\frac{{15}}{9}\,$  $\Rightarrow$ which is not possible $\Rightarrow \theta\ = 0 or \pi /2$ 
but $\theta \pi /2 \Rightarrow \theta = 0$
Hence $P (5, 0)$ i.e. end of major axis