the equation of the common tangent touching the circle (x-3)^2+y^2=9 and the parabola y^2=4x above the x-axis is

Center of circle: (3,0)

Equation of tangent on parabola:

y=mx+a/m

a=1

y=mx+1/m

Perpendicular Distance from the center of the circle = 3

|3m+1/m|/(1+m)^1/2=3

m²=1/3 => m=+1/√3 or -1/√3

As tangent is above x-axis, slope is (+)ve

Therefore, Equation of tangent:

y=x/√3+√3
=> √3y=x+3

Dear Student,
Please find below the solution to your problem.

Center of circle: (3,0)
Equation of tangent on parabola:
y=mx+a/m
a=1
y=mx+1/m
Perpendicular Distance from the center of the circle = 3
|3m+1/m|/(1+m)1/2=3
m2=1/3 => m=+1/√3 or -1/√3
As tangent is above x-axis, slope is (+)ve
Therefore,
Equation of tangent:
y=x/√3+√3
=>√3y=x+3

Thanks and Regards