JEE Class main Answered

Let A is centre of small circle of radius 3 cm . Let B is centre of circle of radius 6 cm .

Let C be the centre of circle that just encloses the circles of radii 3cm and 6 cm as shown in figure.

Common tangent KL make contact with circles at D and E. 

Hence if we draw lines AD and BE , these lines are  perpendicular to KL.

Let us draw a perpendicular line CF from C to KL.

In ΔXBE, we have AD | | BE . Hence we have

From above expression, we get ,  XA = AB = 9 cm

In ΔXCF, we have AD | | CF . Hence we have

Hence we get , CF = (5/3) AD = (5/3) × 3 = 5 cm

Since CF is perpendicular to the chord KL , we get

KL = 2 KF = 15 cm