A straight line AB is divided at C so that AC=3CB. Circles are described on AC and CB as diameters and a common tangent meets AB produced at B. Show that BD is equal to the radius of the smaller circle.
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Solution
We take the co-ordinates of A and B as (−a,0) and (a,0) respectively so that the mid-point of AB is the origin and AB the x-axis. If (h,0) are the co-ordinates of C, then since AC=3CB, we have h=3.a+1.(−a)3+1=12a. On AC and CB as diameters we draw two circles so that they touch each other at C(12a,0). Their centres are P(−14a,0) and Q(34a,0) and radii are 34a and 14a respectively. Let a common tangent RS meet AB produced in D. Let (α,0) be the co-ordinates of D. Then since D divides PQ externally in the ratio of the radii 34a:14a i.e. 3:1, we have α=3.(34a)−1.(−14a)3−1=54a. ∴OD=54a and OB=a so that BD=5$a−a=14a. Hence BD is equal to the radius of the smaller circle.