Area of circle inscribed in two quarter circles So our math teacher gave us this puzzle:

We need to find the area of the circle.
The equation for the two quarter circles are:
$x^2+y^2=1$,
$(x-1)^2+y^2=1$.
I found that the equation for the circle is $(x-0.5)^2+(y-a)^2=a^2$.
I want to find out $a$, as the circle is tangent to both the two quarter circles.
Thanks!
 A: Here's your diagram with a few lines, points and line lengths added:

In particular, $A$ is the origin, $O$ is the center of the green circle, $B$ is the point where the green and red circles are tangent to each other, $C$ is the tangent point between the green circle and the $x$-axis (so it's also where the line normal to the $x$-axis starts from to go through $O$), and $DF$ (which goes through $B$) is the common tangent line at $B$ to the green and red circles. Thus, since $OB$ and $AB$ are both perpendicular to $DF$ at $B$, these lines coincide, i.e., $AB$ is the common normal line to $DF$. Thus, it goes through the centers of both circles, i.e., both $A$ and $O$.
Since $\lvert AB\rvert = 1$, with $\lvert OB\rvert = a$, then $\lvert OA\rvert = \lvert AB\rvert - \lvert OB\rvert = 1 - a$. Also, $\lvert OC\rvert = a$, and $\lvert AC\rvert = 0.5$ due to symmetry (as you've already determined in your circle equation). Since $\triangle ACO$ is right-angled, using the Pythagorean theorem gives
$$\begin{equation}\begin{aligned}
\lvert AC\rvert^2 + \lvert OC\rvert^2 & = \lvert OA\rvert^2 \\
0.5^2 + a^2 & = (1-a)^2 \\
0.25 + a^2 & = 1 - 2a + a^2 \\
2a & = 0.75 \\
a & = 0.375
\end{aligned}\end{equation}$$
