Hello, Insaeno!
This is not a simple problem . . .
Construct a triangle ABC, having given $\displaystyle \angle ACB = 90^o$, the hypotenuse $\displaystyle AB = 6.4$ cm,
and the sum of the remaining sides $\displaystyle AC$ and $\displaystyle BC$ is 8 cm.
Since right triangle $\displaystyle ACB$ can be inscribed in a semicircle,
we have this diagram: Code:

* * * C
*  o*
*  o o *
* o o *
o  o
* o  o*
 A o     +     o B 
3.2 3.2
The circle has the equation: .$\displaystyle x^2 + y^2 \:=\:3.2^2$ . [1]
The locus of point $\displaystyle C$ where $\displaystyle AC + BC \,= \,8$ is an ellipse.
. . We have: .$\displaystyle a = 4$ and $\displaystyle c = 3.2$
Then: .$\displaystyle b^2 \:=\:a^2c^2\:=\:4^2  3.2^2 \:=\:5.76$
. . The equation of the ellipse is: .$\displaystyle \frac{x^2}{16} + \frac{y^2}{5.76} \:=\:1$ . [2]
And solve the system of equations.
I got: .$\displaystyle x \,= \,\pm\sqrt{7},\;\;y \,= \,\pm1.8$