MHB Find the area of the shaded region

[anemone]

Points $$P$$ and $$Q$$ are centers of the circles as shown below. Chord $$AB$$ is tangent to the circle with center $$P$$. Given that the line $$PQ$$ is parallel to chord $$AB$$ and $$AB=x$$ units, find the area of the shaded region in yellow.

[TIKZ]
\draw [<->] (0.2,1) -- (5.8, 1);
\begin{scope}
\draw (3,0) circle(3);
\end{scope}
\begin{scope}
\draw (1,0) circle(1);
\end{scope}
\coordinate[label=above: P] (P) at (1,0);
\coordinate[label=above: Q] (Q) at (3,0);
\coordinate[label=left: A] (A) at (0.2,1);
\coordinate[label=right: B] (B) at (5.8,1);
\coordinate[label=above: $x$] (x) at (2.8,1);
\filldraw (1,0) circle (2pt);
\filldraw (3,0) circle (2pt);

\draw[fill=red,fill opacity=0.6] (1,0) circle (1);
\draw[fill=yellow,fill opacity=0.5] (3,0) circle (3);
[/TIKZ]

I am not seeking for help, I just want to share an interesting geometry problem at MHB with the hope our members enjoy it. :)

 

[anemone]

For those who are interested, you are welcome to post your solution here! I initially wanted to post this in the Challenges subforum, but was afraid this might not difficult enough to be claimed as a challenge problem. (Smile)

 

[anemone]

Solution of MarkFL:

Spoiler

[TIKZ]
\draw [<->] (0.2,1) -- (5.8, 1);
\begin{scope}
\draw (3,0) circle(3);
\end{scope}
\begin{scope}
\draw (1,0) circle(1);
\end{scope}
\coordinate[label=below: P] (P) at (1,0);
\coordinate[label=below: Q] (Q) at (3,0);
\coordinate[label=left: A] (A) at (0.2,1);
\coordinate[label=right: B] (B) at (5.8,1);
\coordinate (T) at (3,1);
\coordinate[label=above: $\frac x2$] (x) at (1.9,1);
\coordinate[label=above: $R$] (R) at (2.2,0.0001);
\coordinate[label=above: $r$] (r) at (3.2,0.1);
\filldraw (1,0) circle (2pt);
\filldraw (3,0) circle (2pt);
\draw (Q)-- (A);
\draw (Q)-- (T);
\draw[fill=red,fill opacity=0.6] (1,0) circle (1);
\draw[fill=yellow,fill opacity=0.5] (3,0) circle (3);
\draw (3,0.9) rectangle +(-0.1, 0.1);
[/TIKZ]
Let the radius of the smaller circle be $$r$$ and the radius of the bigger circle be $$R$$.

Next, build a right-angled triangle where the base of it is half of $$AB$$, i.e. $$\dfrac{x}{2}$$.

Now, by applying the Pythagoras' theorem, we get

$$\begin{align*}\text{Area of shaded region}&=\pi(R^2-r^2)\\&=\pi\left(\dfrac{x}{2}\right)^2\\&=\dfrac{\pi x^2}{4}\end{align*}$$