TITLE: area of figure in sector of intersecting circles
QUESTION [6 upvotes]: I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.

|AB| = a, P is midpoint of segment AB

REPLY [3 votes]: Notes

In the comment, I have mentioned the hardest part of the problem is figuring out $r$,
the radius of the inscribed circle.   
In John's answer, he questioned whether we can derive $r$ using inversion.
Indeed, we can.
Since Martin's answer already covered everything after you get $r$. I'll limit
this answer to the derivation of $r$ using inversion.

Derivation
Choose a new coordinate system such that $P$ is the origin $(0,0)$ and the two points
$A, B$ are located at $(-1,0)$ and $(1,0)$ respectively.
Perform an inversion of the plane at $P$ with respect to the unit circle $x^2+y^2  = 1$.
$$\mathbb{R}^2 \ni (x,y) \quad\mapsto\quad  \left( \frac{x}{x^2 + y^2}, \frac{y}{x^2 + y^2} \right) \in \mathbb{R}^2$$

The points $A, B$ mapped back to themselves.
The two small circles with diameter $AP$ and $PB$ mapped to two straight lines
$$\begin{cases}
\ell_A &= \{\; (x,y) \in \mathbb{R}^2 : x = -1 \;\}\\
\ell_B &= \{\; (x,y) \in \mathbb{R}^2 : x = +1 \;\}
\end{cases}$$
The two big circles centered at $A, B$ passing through $C$ mapped to two
circles of radius $\frac23$ centered at $X = (-\frac13, 0)$ and $Y = (\frac13,0)$ respectively.
$$\begin{cases}
c_X &= \{ \; (x,y) \in \mathbb{R}^2 : (x+\frac13)^2 + y^2 = (\frac23)^2 \;\}\\
c_Y &= \{ \; (x,y) \in \mathbb{R}^2 : (x-\frac13)^2 + y^2 = (\frac23)^2 \;\}
\end{cases}$$
The inscribed circle get mapped to a circle $c_0$ which is kissing the two lines $\ell_A, \ell_B$ and two circles $c_X, c_Y$. Since $c_0$ is sandwiched between $\ell_A, \ell_B$, its radius is clearly $1$. 

Let $Z_0 = (0,h_0)$ be the center of $c_0$, we have
$$XZ_0 = 1 + \frac23 = \frac53\quad\implies\quad
h_0 = \sqrt{\left(\frac53\right)^2 - \left(\frac13\right)^2} = \frac{\sqrt{24}}{3}$$
This implies $c_0$ intersect the $y$-axis at 
$$
\left(0, \frac{\sqrt{24}}{3}-1\right)
\quad\text{ and }\quad
\left(0,\frac{\sqrt{24}}{3}+1\right)
$$
and hence the original inscribed circle intersect $y$-axis at
$$
\left(0,\frac{1}{\frac{\sqrt{24}}{3}-1}\right)
\quad\text{ and }\quad
\left(0,\frac{1}{\frac{\sqrt{24}}{3}+1}\right)
$$
As a result, the radius of the inscribed circle is given by
$$
\frac{r_0}{AB} = 
\frac14\left(\frac{1}{\frac{\sqrt{24}}{3}-1} - \frac{1}{\frac{\sqrt{24}}{3}+1}\right)
= \frac{3}{10}
$$
matching what I described in comment.
Bonus
This approach using inversion is slightly longer than one working with the circles directly.
However, it leads to an interesting extension!
Since $\ell_A$ and $\ell_B$ are parallel, we
can stack on top of $c_0$ an infinite sequences of circles $c_1, c_2, \ldots$ of radius $1$.
The $n^{th}$ circle will be centered at
$$Z_n = (0,h_n) = \left(0, \frac{\sqrt{24}}{3} + 2n\right)$$
If we invert these circle back, we will obtain a sequence of circles inserted between
the inscribed circle and the two small circles. The radius of the $n^{th}$ circle
will be given by the formula
$$\frac{r_n}{AB} = \frac14
\left(\frac{1}{\frac{\sqrt{24}}{3}+2n-1} - \frac{1}{\frac{\sqrt{24}}{3}+2n+1}\right)
= \frac{3}{10 + 16\sqrt{6}n+24n^2}
$$
This is the sequence of circles mentioned in Martin's extended version of this problem.