TITLE: 3 circles internal tangent
QUESTION [9 upvotes]: My friend show me the diagram above , and ask me 
"What is the area of a BLACK circle with radius of 1 of BLUE circle?"
So, I solved it by algebraic method.
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Let center of  $\color{black}{BLACK}$ circle be $(0,0)$.
We can set, 
$x^2 + (y-R)^2 = R^2$ , where $R$ means radius of $\color{red}{RED}$ circle.
$(x-p)^2 + (y-r)^2 = r^2 $,  where $(p,r)$ means center of $\color{blue}{BLUE}$ circle.
$$$$
These can imply
$    2R=r+ \sqrt{p^2 + r^2}$
$p^2 + (R-r)^2 = (R+r)^2 $
So, 
$ 2r=R$
$$$$
But he wants not algebraic but Geometrical Method.
How can I show  $ 2r=R$ with Geometrical Method?
Really thank you.
$$$$
(Actually I constructed the diagram with algebraic methed, 
but I'd like to know how construct this whit Geometrical method.)

REPLY [2 votes]: \begin{align}
\triangle ADB:\quad
|DB|^2&=
(\tfrac{R}2+r)^2
-(\tfrac{R}2-r)^2
=2rR
,\\
\triangle BDO:\quad
|DB|^2&=
(R-r)^2-r^2
=R(R-2r)
.
\end{align}
Hence,
\begin{align} 
R(R-2r)&=2rR
,\\
R&=4r
.
\end{align}