What is the length of $AD$ if distance between centres of the two circles is $9 cm$?

 

Concept: A chord is a line segment that connects any two points on a circle's circumference. The perpendicular from the centre of a circle bisects the chord.

As since $O_{1}B$ and $O_2\mathit{B}$ are perpendicular from centre to the chord,  $O_1{O}_{2}CB$ is a rectangle.

$\Rightarrow O_1{O}_2=BC=9 cm$         ….(i)

Also, the perpendicular from centre to a chord bisects the chord.

$AB=BE$      …. (ii) 

and 

$CE=CD$       ….(iii) 

Now, 

$AD=AB+BE+CE+CD$  

From equation (ii) and (iii),

$\Rightarrow AD=2BE+2CE$

  $\Rightarrow AD=2(BE+CE)$

$\Rightarrow AD=2BC$ 

From equation (i),

$\Rightarrow AD=2\times 9=18 cm$

Hence, the length of  $AD$ is  18 cm.