area of figure in sector of intersecting circles 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

I believe we can use Descartes Circle Theorem.  We can identify 4 circles:


*

*the circle $O_A$ with center at $A$

*circle with diameter $\overline{AP}$

*circle with diameter $\overline{PB}$

*unknown circle 


Unfortunately, these 4 circles are not mutually tangent since the circle with diameter $\overline{AP}$ is not tangent to the circle $O_A$ with center at $A$.

The general Apollonius problem involves finding a circle tangent to 3 other circles (regardless of mutual tangency).
One, rather boring, solution is to solve 3 quadratic equations.  Let $x,y,r$ be the center and radius of the unknown circle.
$$  (x-x_k)^2 + (y-y_k)^2 = (r \pm r_k)^2$$
where $k = 1,2,3$.  Using the diagram, we put in the centers and radii.
$$ x^2 + y^2 = (r \pm 4)^2$$
$$ (x-1)^2 + y^2 = (r \pm 1)^2$$
$$ (x-3)^2 + y^2 = (r \pm 1)^2$$
In order to get the signs right, our circle is inside the first circle and outside the other two.
$$ x^2 + y^2 = (r - 4)^2$$
$$ (x-1)^2 + y^2 = (r + 1)^2$$
$$ (x-3)^2 + y^2 = (r + 1)^2$$
The fact that $x = 2$ should have been clear from the diagram.  Writing the system again:
$$ 2^2 + y^2 = (r - 4)^2$$
$$ 1^2 + y^2 = (r + 1)^2$$
We can subtract the first and second equations, it is possible to solve the linear equation for $r$:
$$ 3 = 2^2 - 1^2 = (r-4)^2 - (r+1)^2 = (r^2-8r + 16) - (r^2 + 2r + 1) = -10r + 15$$
The answer is $r = \tfrac{6}{5}$.  If we rescale so the radius with center $0$ is the unit circle, we get $\boxed{r=\tfrac{3}{10}}$

A more interesting solution would exploit the symmetry of the problem or use inversion somehow.  
