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