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

Any tangent to y² = 4x is of the form $\mathrm{y}=\mathrm{mx}+\frac{1}{m}$$\left(\therefore a=1\right)$and this touches the circle

(x-3)²+y²=9  If  $r=d\Rightarrow$ $\left|\frac{m\left(3\right)+\frac{1}{m}-0}{\sqrt{m^2+1}}\right|=3$

Centre of the circle is (3,0) and radius is 3

$\begin{array}{l}\Rightarrow \frac{3m^2+1}{m}=\pm 3\sqrt{m^2+1}\\ \Rightarrow 3m^2+1=\pm 3m\sqrt{m^2+1}\end{array}$

$\Rightarrow 9m^4+1+6m^2=9m^2\left(m^2+1\right)$

$\Rightarrow 9m^4+1+6m^2=9m^2+9m^2$

$3m^2=1\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

If the tangent touches the parabola and circle above the x – axis, then slope m should be positive.

$\therefore m=\frac{1}{\sqrt{3}}$ and the equation is $y=\frac{1}{\sqrt{3}}x+\sqrt{3}or\sqrt{3}y=x+3$