circlesgeometry
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!
Let's get a figure in here, eh?
We can exploit the fact that the points of mutual tangency of the three inner circles (red points in the figure) form an equilateral triangle; we also know that the red points are the midpoints of the segments formed by joining any two of the three blue points (the centers of the inner circles).
We then deduce that the triangle formed by the red points is half the scale of of the triangle formed by the blue points, and find that the equilateral triangle formed by the blue points has a side length of 6 ft., and that the inner circles have a radius of 3 ft. Using the law of cosines, we reckon that the distance from a blue point to the center of the triangle formed by the blue points is $2\sqrt{3}$ ft. Adding to that the radius of an inner circle, we find that the radius of the outer circle is $3+2\sqrt{3}$ ft.
The area of the outer circle is $\approx$ 131.27 square feet.
In the midst of a competition you can make all sorts of careless mistakes, but you should at least now see your answer cannot be correct, since $16<8\pi$ so your answer is negative, which is obviously undesirable.
When you get a negative area, what happened is that you took away too much; in this case that is true; you took away the white region twice while only adding in the square once.
The error is relatively easy to fix, fortunately. Your construction also took away the shaded region once (one part for each quarter-circle), so if you add back in the entire square then you have
which means you have added exactly the shaded region, exactly once; that is the desired area. This does agree with the $32-2\pi(16)/4 = 32-8\pi$ of Sry's comment. (As a sanity check, since $\pi<4$ we have that this answer is at least positive).
Best Answer
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
$$\begin{equation}\begin{aligned} \lvert AC\rvert^2 + \lvert OC\rvert^2 & = \lvert OA\rvert^2 \\ 0.5^2 + a^2 & = (1-a)^2 \\ 0.25 + a^2 & = 1 - 2a + a^2 \\ 2a & = 0.75 \\ a & = 0.375 \end{aligned}\end{equation}$$