Maths

Since both circles touch the x - axis, the point of intersection of the 2 external common tangents must be on the x - axis. Let the point be P(p, 0).
Let centre of circle C be A which is A(5, 2). Radius = 2
Let centre of circle D be B which is B(3,1). Radius = 1.
By similar triangle, PB : PA = 1 : 2.
PB/(PB + BA) = 1/2
2PB = PB + BA
PB = BA
so B is the mid- point of AP
so (p + 5)/2 = 3
p + 5 = 6
p = 1
so the point of intersection is (1,0).