Circle problem in which i wanted to find value of $PQ$ Find a value of $PQ$:  
Let the radius of bigger circle be $R$, and that of the smaller circles be $r_1$ and $r_2$.
Hence we can find value of $R = 10$.
I don't know how to proceed further to find the Value of $r_1$ and $r_2$.
 A: Let $AP =x$, $PQ = y$ and $QB = z$, so by Pythagoras theorem we have $$x+y+z = 20$$
Since $AC = AQ$ and $BP = BC$ we have also: $$x+y=12$$ $$y+z=16$$
so $PQ = y = 8$.

I used a following lemmas:
Lemma 1: $A,H,G$ are collinear.
Proof: Observe a homothety at $G$ wich takes smaller circle to a bigger and let $H'$ be a image of $H$. Then $H'$ is on bigger circle and tangent $CH$ goes to parallel tangent on bigger circle, so it can go only through $A$, so $H'=A$ and thus $A,H,G$ are collinear.
 
Lemma 2: $AF = AC$
Proof: Suppose $A,H,G$ are collinear. Since $BGHD$ is cyclic we can use the power of the point $A$:
 $$ AH \cdot AG = AD\cdot AB$$
If we use the power of the point $A$ with respect to smaller circle we have:
$$ AH\cdot AG = AF^2$$
And if we use the power of the point $A$ with respect to circle $(BCD)$ which is tangent to $AC$ we have: $$AD\cdot AB = AC^2$$
So $AC = AF$.
A: 
Using the power of the point $O$ 
with respect to circle $O_1$ 
we have
\begin{align} 
|OT_2|\cdot|OT_1|&=|OP|^2
,\\
\text{or }\quad
R(R-2r_1)&=(R-(|AD|-r_1))^2
,\\
r_1&=
\sqrt{2R(2R-|AD|)}-(2R-|AD|)
\tag{1}\label{1}
.
\end{align} 
Similarly, 
\begin{align} 
r_2&=\sqrt{2R\cdot|AD|}-|AD|
\tag{2}\label{2}
.
\end{align}
Note that given 
$a=16=4\cdot4$,
$b=12=4\cdot3$ 
means that 
$\triangle ABC$ 
is a scaled version of the famous $3-4-5$ triangle,
so $R$ and $|AD|$ can be easily calculated,
\begin{align} 
R&=4\cdot\frac 52=10
,\\
|AD|&=4\cdot3\cdot\frac 35=\frac{36}5
,\\
r_1&=\frac{16}5
,\\
r_2&=\frac{24}5
,\\
|PQ|&=8
.
\end{align}
A: I am using the lemmas provided by @greedoid . Then the calculation of PQ can be simplified as:-
$PQ = AQ + BP - AB = 12 + 16 - 20 = 8$
