Answer:
Option C
Explanation:
Any tangent to parabola y2 =8x is $y= mx+\frac{2}{m}$ .............(i)
It touches the circle $x^{2}+y^{2}-12x+4=0$
if the length of perpendicular from the centre (6,0) is equal to radis $\sqrt{32}$
$\therefore$ $\frac{6m+\frac{2}{m}}{\sqrt{m^{2}+1}}=\pm\sqrt{32}$
$\Rightarrow \left(3m+\frac{1}{m}\right)^{2}=8(m^{2}+1)$
$\Rightarrow(3m^{2}+1)^{2}=8(m^{4}+m^{2})$
$\Rightarrow m^{4}-2m^{2}+1=0\Rightarrow m=\pm1$
Hence, the required tangents are y=x+2 and y=-x-2
