small circle area I have tried to solve this problem, but I can not see its solution.
It is to calculate the area of the small circle, which is tangent to the left arc to the right and to the Y axis in points that I do not know.
If I knew these points I could easily reach the solution, some idea of how to approach it
attached image of the problem
 A: Let $(a,b)$ the center of the circle and $r$ the radius of the larger circles. Notice that $a$ is also the radius of the circle in question.  We have
$$r-\text{distance from $(a,b)$ to $(0,r)$}=a$$
and
$$\text{distance from $(a,b)$ to $(r,0)$}-r=a,$$
that is
$$r-\sqrt{a^2+(b-r)^2}=a\quad\text{and}\quad
\sqrt{(a-r)^2+b^2}-r=a.$$
From here we easily deduce $b=\frac43r$ and $a=\frac49r$.
A: Use the fact that the point of tangency is on the line between the centers.  Let the point of tangency be $(x,y)$ and the radius of the small circle be $r$ with the center of the small circle at $(r,z)$.  You have that $(x,y)$ is on both circles, that the slopes from the centers to $(x,y)$ are equal and that the segment is divided in the ratio of the radii.  This gives four equations in four unknowns.
$$(9-x)^2+y^2=81\\(x-r)^2+(z-y)^2=r^2\\\frac {x-r}{9-r}=\frac r{9+r}\\\frac{z-y}z=\frac r{9+r}$$
 Alpha seems to say I am still an equation short.  
The missing equation reflects the tangency of the small circle with the large circle on the left.  As the line from $(0,9)$ through $(r,z)$ hits the point of tangency we have $$r^2+(z-9)^2=(9-r)^2$$
Now Alpha finds 
$$r=4, x=\frac {72,13},y=\frac {108}{13},z=12$$
A: Label as indicated on the graph:
$\hspace{1.5cm}$$
The equations:
$$\begin{align}(x-r)^2+(y-s)^2&=r^2 \ \ (\text{small circle})\\
(a-r)^2+(b-s)^2&=r^2 \ \ (\text{small circle})\\
(x-9)^2+(y-0)^2&=9^2 \ \ (\text{horizontal semicircle})\\
(a-9)^2+(b-0)^2&=9^2 \ \ (\text{vertical semicircle})\\
s^2+(9-r)^2&=(9+r)^2 \ \ (\text{large triangle})\\
(s-9)^2&=(9-2r)\cdot 9 \ \ (\text{green line})\\
\end{align}$$
We will (set up and make WA) solve the $6$ equations with $6$ unknowns and get:
$$a=\frac{36}{5}; b=\frac{72}{5}; r=4; s=12; x=\frac{72}{13}; y=\frac{108}{13}.$$
Hence, the area is: $A=\pi\cdot r^2=16\pi \approx 50.265$.
