Consider circles $C_1$ and $C_2$ touching both the axes and passing through $\left(4,\mathrm{}4\right),$ then the $x$-intercept of the common chord of the circles is

Since the circle touches both the axes and passes through $\left(4,\mathrm{}4\right)$, it lies in the first quadrant, so its equation is
${\left(x-r\right)}^2+{\left(y-r\right)}^2=r^2$
$\left(4,\mathrm{}4\right)$ lies on it, hence
$r^2-16r+32\Rightarrow r_1+r_2=16$
$C_1$ is ${\left(x-r_1\right)}^2+{\left(y-r_1\right)}^2=r_1^2$
$C_2$ is ${\left(x-r_2\right)}^2+{\left(y-r_2\right)}^2=r_2^2$
Equation of the common chord is
$\left(2x-\left(r_1+r_2\right)\right)+\left(2y-\left(r_1+r_2\right)\right)+\left(r_1+r_2\right)=0$
$\Rightarrow x+y=8$

Hence, the $x$-intercept is $8$