If $$a>2b>0$$    then the positive value of $$m$$ for which $$y = mx - b\sqrt {1 + {m^2}} $$     is a common tangent to $${x^2} + {y^2} = {b^2}$$   and $${\left( {x - a} \right)^2} + {y^2} = {b^2}$$    is :

Solution :
Given that $$a>2b>0$$    and $$m>0$$
Also $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = mx - b\sqrt {1 + {m^2}} .....(1)$$
is tangent to $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + {y^2} = {b^2}.....(2)$$
as well as to $${\left( {x - a} \right)^2} + {y^2} = {b^2}.....(3)$$
$$\because $$ (1) is tangent to (3)
$$\therefore \left| {\frac{{am - b\sqrt {1 + {m^2}} }}{{\sqrt {{m^2} + 1} }}} \right| = b$$
[ length of perpendicular from $$\left( {a,\,0} \right)$$  to (1) $$=$$ radius $$b$$ ]
$$\eqalign{ & \Rightarrow am - b\sqrt {1 + {m^2}} = \pm b\sqrt {1 + {m^2}} \cr & \Rightarrow am - 2b\sqrt {1 + {m^2}} = 0 \cr & {\text{or, }}am = 0\,\,\left( {{\text{not possible as }}a,\,m > 0} \right) \cr & \Rightarrow {a^2}{m^2} = 4{b^2}\left( {1 + {m^2}} \right) \cr & \Rightarrow {m^2} = \frac{{4{b^2}}}{{{a^2} - 4{b^2}}} \cr & \Rightarrow m = \frac{{2b}}{{\sqrt {{a^2} - 4{b^2}} }}\,\,\,\left( {\because m > 0} \right) \cr} $$