The equation of the common tangent touching the circle $${\left( {x - 3} \right)^2} + {y^2} = 9$$    and the parabola $${y^2} = 4x$$  above the $$x$$-axis is-

Solution :
Let the equation of tangent to $${y^2} = 4x$$   be $$y = mx + \frac{1}{m}$$   where $$m$$ is the slope of the tangent.
If it is tangent to the circle $${\left( {x - 3} \right)^2} + {y^2} = 9$$     then length of perpendicular to tangent from centre $$\left( {3,\,0} \right)$$  should be equal to the radius $$3.$$
$$\eqalign{ & \therefore \frac{{3m + \frac{1}{m}}}{{\sqrt {{m^2} + 1} }} = 3 \cr & \Rightarrow 9{m^2} + \frac{1}{{{m^2}}} + 6 = 9{m^2} + 9 \cr & \Rightarrow m = \pm \frac{1}{{\sqrt 3 }} \cr} $$
$$\therefore $$ Tangents are $$x - y\sqrt 3 + 3 = 0$$    and $$x + y\sqrt 3 + 3 = 0$$    out of which $$x - y\sqrt 3 + 3 = 0$$    meets the parabola at $$\left( {3,\,2\sqrt 3 } \right)$$   i.e., above $$x$$-axis.