2 circles of different radii R and r touch each other externally..the three common tangents form a triangle.show that area of the triangle is [2(rR)^(3/2)   /    R-r]

Let O1 and O2 be centers of circles with radii r and R resp. We take the common point as origin (O) and line through centers as y-axis, Thus x-axis will be third tangent. O1 = (0,r) and O2(0,-R). Let P =(0,y) be point of intersection of tangents. Let these tangents intersect x-axis at Q (x,0) and R (-x,0). If we can find x and y then area of triangle is (1/2)*(2x)*y

 

For y we can say

(y-r)/(y+R)=(r/R) solve for y from here.

From trignometry sin(angle OPQ) = r/y

So tan( angle OPQ) = r/sqrt(r^2+y^2) = x/y

From here solve for x and hence area of triangle is xy