Find $DF$ in a triangle $DEF$ Consider we have a triangle $ABC$ where there are three points $D$, $E$ & $F$ such as point $D$ lies on the segment $AE$, point $E$ lies on $BF$, point $F$ lies on $CD$. We also know that center of a circle over ABC is also a center of a circle inside $DEF$. $DFE$ angle is $90^\circ$, $DE/EF = 5/3$, radius of circle around $ABC$ is $14$ and $S$ (area of $ABC$), K (area of DEF), $S/K=9.8$. I need to find $DF$. Help me please, I'd be very grateful if you could do it as fast as you can. Sorry for inconvenience.

I refer to the diagram of Victor Liu's answer. This is an analytical
verification that $DF=8$ (which means that $u=2$) but omits some details. Using the equation of the circle centered at $(0,0)$ with radius $14$, the equation of $AE$ (tangent to the small circle at the point $(x,y)=(-8/5,6/5)$) 
$$
y=\frac{4}{3}\left( x+\frac{8}{5}\right) +\frac{6}{5}
$$
and the equations of $BF$ ($y=-2)$ and $CD$ ($x=2$), we get the coordinates
of the vertices of triangle $ABC$:
$$
A\left( -\frac{8}{5}+\frac{24}{5}\sqrt{3},\frac{32}{5}\sqrt{3}+\frac{6}{5}
\right) ,\qquad B(-8\sqrt{3},-2),\qquad C(2,-8\sqrt{3}).
$$
The coordinates of the vertices of the right triangle's $DEF$ are
$$
D(2,8),\qquad E(-4,-2),\qquad F(2,-2).
$$
The lengths of the sides of $ABC$ computed by the distance formula are
$$
a =BC=14\sqrt{2}, \qquad
b =AC=\frac{42}{5}\sqrt{10},
\qquad
c =AB=\frac{56}{5}
\sqrt{5}.
$$
The semi-perimeter $p$ of $ABC$ is thus
$$
p=\frac{a+b+c}{2}=7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}.
$$
By Heron's formula the area of $ABC$ is 
$$
S=S_{ABC}=\sqrt{p(p-a)(p-b)(p-c)}.
$$
Since 
$$
\begin{eqnarray*}
&&p(p-a)(p-b)(p-c) \\
&=&\left( 7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right) \left(
-7\sqrt{2}+\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right)  \\
&&\times \left( 7\sqrt{2}-\frac{21}{5}\sqrt{10}+\frac{28}{5}\sqrt{5}\right)
\left( 7\sqrt{2}+\frac{21}{5}\sqrt{10}-\frac{28}{5}\sqrt{5}\right)  \\
&=&\frac{1382976}{25},
\end{eqnarray*}
$$
we get
$$
S=\sqrt{\frac{1382976}{25}}=\frac{1176}{5}.
$$
The area of $DEF$ is $$K=S_{DEF}=\frac{EF\times DF}{2}=\frac{6\times 8}{2}=24$$
and the ratio $$\frac{S}{K}=\frac{1176/5}{24}=\frac{49}{5}=9.8,$$ as given.

Added: The scale is uniform throughout the following diagram drawn with the calculated equations

