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Question

A straight line AB is divided at C so that AC=3 CB. 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=3 CB, 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)31=54a.
OD=54a and OB=a so that
BD=5$aa=14a.
Hence BD is equal to the radius of the smaller circle.
923187_1007404_ans_9c2f4a2df2b94a5c8c1e4c37e254725f.jpg

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