Answer:
[tex]\displaystyle {\rm AZ} = \frac{1}{2}\, ({\rm AB}) = 4[/tex].
Step-by-step explanation:
Refer to the diagram attached. Let [tex]{\rm D[/tex] denote the center of the circle which is tangent to [tex]{\rm AB}[/tex] at [tex]{\rm A}[/tex]. Let [tex]{\rm E}[/tex] denote the center of the circle which is tangent to [tex]{\rm AB}[/tex] at [tex]{\rm B}[/tex].
Because [tex]{\rm AB}[/tex] is tangent to the circle centered at [tex]{\rm D}[/tex], the tangent line [tex]{\rm AB}[/tex] would be perpendicular to [tex]{\rm DA}[/tex], which joins the center of the circle (point [tex]{\rm D}[/tex]) and the point at which the tangent line touches the circle (point [tex]{\rm A}[/tex].) Hence, [tex]\angle {\rm DAZ} = \angle {\rm DAB} = 90^{\circ}[/tex]. Similarly, [tex]\angle {\rm DCZ} = 90^{\circ}[/tex]. The same property applies to the circle centered at [tex]{\rm E}[/tex], such that [tex]\angle {\rm ECZ} = 90^{\circ}[/tex] and [tex]\angle {\rm EBZ} = \angle {\rm EBA} = 90^{\circ}[/tex].
Hence, [tex]\triangle {\rm DAZ}[/tex], [tex]\triangle {\rm DCZ}[/tex], [tex]\triangle {\rm ECZ}[/tex], and [tex]\triangle {\rm EBZ}[/tex] are right triangles.
Observe that in right triangles [tex]RT \triangle {\rm DAZ}[/tex] and [tex]RT \triangle {\rm DCZ}[/tex]:
Hence, these two right triangles are congruent ([tex]RT \triangle {\rm DAZ} \cong RT \triangle {\rm DCZ}[/tex]) by "HL" height-hypotenuse congruence- a special property of right triangles. Because these two triangles are congruent, length of [tex]{\rm AZ}[/tex] would be the same as that of [tex]{\rm CZ}[/tex] (such that [tex]{\rm AZ} = {\rm CZ}[/tex].)
Similarly, [tex]RT \triangle {\rm ECZ} \cong \triangle {\rm EBZ}[/tex], such that [tex]{\rm CZ} = {\rm BZ}[/tex].
In other words, [tex]{\rm AZ} = {\rm CZ} = {\rm BZ}[/tex]. Moreover, since [tex]{\rm Z}[/tex] is a point on line [tex]{\rm AB}[/tex], [tex]{\rm AZ} + {\rm BZ} = {\rm AB}[/tex]. Hence:
[tex]{\rm AZ} + {\rm AZ} = {\rm AZ} + {\rm BZ} = {\rm AB}[/tex].
[tex]2\, ({\rm AZ}) = {\rm AB}[/tex].
[tex]\displaystyle {\rm AZ} = \frac{{\rm AB}}{2}[/tex].
In other words, regardless of the radius of the two circles, length of [tex]{\rm AZ}[/tex] would be half of that of [tex]{\rm AB}[/tex].
