Equation of a common tangent to the circle, $${x^2} + {y^2} - 6x = 0$$    and the parabola, $${y^2} = 4x,$$  is :

Solution :
Since, the equation of tangent to parabola $${y^2} = 4x$$   is
$$y = mx + \frac{1}{m}.....(1)$$
The line (a) is also the tangent to circle
$${x^2} + {y^2} - 6x = 0$$
Then centre of circle $$ = \left( {3,\,0} \right)$$
radius of circle $$= 3$$
The perpendicular distance from centre to tangent is equal to the radius of circle
$$\eqalign{ & \therefore \frac{{\left| {3m + \frac{1}{m}} \right|}}{{\sqrt {1 + {m^2}} }} = 3 \cr & \Rightarrow {\left( {3m + \frac{1}{m}} \right)^2} = 9\left( {1 + {m^2}} \right) \cr & \Rightarrow m = \pm \frac{1}{{\sqrt 3 }} \cr} $$
Then, from equation (1) : $$y = \pm \frac{1}{{\sqrt 3 }}x \pm \sqrt 3 $$
Hence, $$\sqrt 3 y = x + 3$$   is one of the required common tangent.