A line is a common tangent to the circle ${(x-3)}^2+y^2=9$  and the parabola $y^2=4x.$  If the two points of contact $(a,b)\hspace{0.17em}and\hspace{0.17em}(c,d)$  are distinct and lie in the first quadrant, then $2(a+c)$  is equal to ___.

Let on the parabola $(a,b)\equiv (t^2,2t)\equiv A$ 
Then equation of tangent to parabola at A is   $ty=x+t^2$
This line touches the circle also.
$\begin{array}{l}\therefore \left|\frac{3-0+t^2}{\sqrt{1+t^2}}\right|=3\\ \Rightarrow {(3+t^2)}^2=9(1+t^2)\\ \Rightarrow t=0,-\sqrt{3},\sqrt{3}\end{array}$
So, $A\equiv (3,2\sqrt{3})$  and equation of tangent is  $\sqrt{3}y=x+3.$
Equation of normal to circle perpendicular to the above tangent is  $y=-\sqrt{3}x+3\sqrt{3}.$
Solving above lines we get point of contact on the circle as  $(\frac{3}{2},\frac{3\sqrt{3}}{2})$.
$\Rightarrow 2(a+c)=2(3+\frac{3}{2})=9$