The focal chord of y² = 16x touches the circle (x–6)² + y² = 2. Then the positive slope of the chord is

Parabola is y² =16x$\Rightarrow$focusS(4,0) ,
EQ. of focal chord is y=mx+c it passes through (4,0) then c=-4m

and chord is y= m (x-4) ,

It touches the circle then r=d

$$\begin{gathered} \Rightarrow \sqrt{2}=\left|\frac{6m-4m}{\sqrt{m^2+1}}\right| \\ \Rightarrow 2(m^2+1)=4m^2 \end{gathered}$$

$\Rightarrow {\mathrm{m}}^2\equiv 1\Rightarrow \mathrm{m}=1$