The focal chord of $y^2=16x$ is tangent to $(x-6{)}^2+y^2=2$, then possible values of the slope of this chord are

The correct option is A $\{-1,1\}$

For a line $y=mx+c$ to be tangent to $(x-6{)}^2+y^2=2$

$\sqrt{2}=\frac{6m+c}{\sqrt{m^2+1}}$

Focal chord passes through focus hence, passes through $(4,0)$

$⟹c=-4m$

$⟹\sqrt{2}=\frac{2m}{\sqrt{m^2+1}}$

squaring on both sides

$2m^2+2=4m^2$

$m=\pm 1$