Official solution
We know,
Equation of tangent to the parabola y2 = 4ax is,
y = mx + \[{a \over m}\]
\[ \therefore \] Equation of tangent to the parabola y2 = 4x is,
y = mx + \[{1 \over m}\]
\[ \Rightarrow \] m2x - ym + 1 = 0
This tangent is also the tangent to the circle x2 + y2 - 6x = 0
So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.
Here center is at (3, 0) of the circle and radius = 3
\[ \therefore \] \[\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3\]
\[ \Rightarrow \] (3m2 + 1)2 = 9(m4 + m2)
\[ \Rightarrow \] 9m4 + 6m2 + 1 = 9m4 + 9m2
\[ \Rightarrow \] 3m2 = 1
\[ \Rightarrow \] m = \[ \pm \] \[{1 \over {\sqrt 3 }}\]
So, possible tangents are
y = \[{1 \over {\sqrt 3 }}\]x + \[\sqrt 3 \]
\[ \Rightarrow \] \[\sqrt 3 \]y = x + 3
or y = - \[{x \over {\sqrt 3 }}\] - \[\sqrt 3 \]
\[ \Rightarrow \] \[\sqrt 3 y\] = - x - 3