Sorry for bad drawing. In order to find $PQ$ I draw radiuses $r_1$ and $r_2$ and according to image $PQ$ is equal to white line which is equal to
$$\sqrt{(r_1+r_2)^2-(r_1-r_2)^2}=2\sqrt{r_1r_2}$$
But I don't know how to find $r_1$ and $r_2$.
Edit:
Here is original image of the problem:
The two circles are not in contact; $PQ \neq 2\sqrt{r_1r_2}\,$ but actually $PQ=r_1+r_2$.
Let's name the blue circle $C$, its center $X$, its tangent point with the large circle $Y$ and the intersection point of $PQ$ and the vertical line $H$.
Let the radius of $C$ be $r$, then $XY=PX=PH=r$.
$$OY=10 \Rightarrow OX=10-r$$ $$OH=10-\frac{36}{5}=\frac{14}{5} \Rightarrow PO=r+\frac{14}{5}$$ $$ PX^2+PO^2=OX^2 \Rightarrow r^2+(r+\frac{14}{5})^2=(10-r)^2 \Rightarrow r=\frac{16}{5} $$
Using a similar method, the radius of the green circle is $\frac{24}{5}$.
Therefore, $PQ=8$.