Radius of a circle touching circle of given size and with given distance from common point to bitangent I'm getting different answers using different methods in this question -

There are 2 spherical balls touching each other at a point P. They both are kept on the ground and the point P is 10 cm above the ground. The radius of one ball is 8 cm. Find the radius of the other ball.

My approach -
Let the radius of the other circle be $r$ cm.
Drawing perpendiculars from the centers of both triangles, as well as from the center of triangle with radius $8$ cm (to other radius), we get a right triangle. Now perpendicular $= r - 8$ cm. Hypotenuse $= r + 8$ cm, and base $= 20 cm$ (Using Tangent Property from same point).
Solving, we get $r = 12.5$
If instead, we use similarity, we get $2$ other answers - $8, \frac{40}{3}$. Indeed, $8$ is wrong.
Please help.


Similar triangles provides the correct answer.
Using (height 1)/(height 2) = (Hypotenuse 1)/(hypotenuse 2), you have (10-8)/(r-8) = 8/(r+8) leading to r = 40/3
The line from P down to the point Q on the ground that meets the ground at right angles has a length of 10 cm, but it not a tangent to either circle, so the base of the larger triangle is not 20cm. 
(Sorry for lack of diagrams, I am new to SE and have just joined this board and haven't worked out how to get diagrams into my answer - I will edit it when I have found out how to do that)

