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We can  construct two circles with equations

x^2 + y^2   =4

(x -3)^2 + y^2  = 1

These circles are tangent at P = (2,0)

We can also construct similar triangles   ABD  and ACE   where  points D and E are  the points of tangency of our line of interest and  the two circles

We can find point  "A"  as follows

Let   a =  the distance from  A  to the smaller circle

AB = (radius of smaller circle + a) = 1 + a

AC = ( radius of larger circle + a)  =   4 + a

BD = 1

CE = 2

So  we  have this relationship  in these similar triangles

AB / AC  =  BD / CE

(1 + a) / ( 4 + a)  =   1/2     cross-multiply

2 (1 + a)  =  1 (4 + a)

2 + 2a  = 4 + a

2a  - a =  4  - 2

a =  2

So  AB  =   1 + 2   = 3

So CA =  (2 + 1 + 3)  = 6

So point A has the coordinates (6,0)

Now since  triangle ABD is right with hypotenuse AB,  then  AD  = sqrt ( AB^2 - BD^2)  = sqrt (3^2 -1^2) =sqrt (8)

The  tangent  of  angle  BAD  = BD / AD  =   1 /sqrt (8)   = the slope of our line

So the equation of our line  going through (6,0) is 

y  =   (1/sqrt (8)) ( x - 6)

Multiply through by  sqrt 8

sqrt(8)y  = x -6

Putting  in standard form

x - sqrt (8) y - 6  = 0

The distance  from point  P  = (2,0)   to this line  is given  by

          abs  [  2 - sqrt (8)(0) - 6 ]               abs (-4)               4

      _______________________  =      _________  =     ___    =   PF

          sqrt  [ 1^2  + [ sqrt (8)]^2                sqrt (9)               3

So   m + n  = 7