Question

two circles with radii a and b touch each other externally. let C be the radius of a circle which touches these two circles as well as a common tangent to the two circles. then prove that 1/root c=1/root a+ 1/root b

Answer 1

Sridhar R seems you are a matric student studying in tenth standard. Diagram you might have got with you. Ok. Just using key board giving explanations will be somehow difficult. Any way let me give the right tips to catch the steps to solve it in an interesting way. 
We have three pairs of circles. First a line parallel to the common tangent and passing through the centre of the third circle with radius c is drawn. That line will have two segments. 
One part will be equal to sq root of (a+c)^2 - (a-c)^2 and 
the other will be equal to sq root of (b+c)^2 - (b-c)^2. Another line parallel to common tangent and passing through the centre of circle, smaller of the first two circles. So this will be exactly equal to the length of the common tangent. Also this will be equal to sq root of (a+b)^2 - (a-b)^2 
So total length = sum of the two segments 
sq root of (a+b)^2 - (a-b)^2 = sq root of (a+c)^2 - (a-c)^2 + sq root of (b+c)^2 - (b-c)^2 
We are so familiar with the identity (a+b)^2 - (a-b)^2 = 4 ab 
So applying this we get sq root (4ab) = sq root (4ac) + sq root (4bc) 
====> sq root (ab) = sq root (ac) + sq root (bc) 
Dividing throughout by sq root of (abc) we get 1/ root(c) = 1/root(b)+ 1/root(a) 

Hope This Helps :)