Find the area of the shaded region

[anemone]

Points \(\displaystyle P\) and \(\displaystyle Q\) are centers of the circles as shown below. Chord \(\displaystyle AB\) is tangent to the circle with center \(\displaystyle P\). Given that the line \(\displaystyle PQ\) is parallel to chord \(\displaystyle AB\) and \(\displaystyle AB=x\) 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 \(\displaystyle r\) and the radius of the bigger circle be \(\displaystyle R\).

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

Now, by applying the Pythagoras' theorem, we get

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