The circle within the sector touches the radii R and the arc.
So what is the area of the inscribed circle?
The answer is actually $$S = \pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$
How can I derive this?
[Math] Find the area of a circle that is inscribed in a circular sector with a radius $R$ and an angle $2x$.
geometrytrigonometry
Related Solutions
Here are some equal angle inscribed angles (they all subtend half the angle of the red arc). It is clear that the area of the last sector is properly contained in the areas of the others.
$\hspace{2.5cm}$
The area of the sector is the sum of the area between the arc and its chord (a constant area) and the area of the triangle, which is $\frac12$ the product of the length of the chord (a constant length) and the altitude of the triangle, which varies as the vertex moves around the circle.
$$A_S=\frac{75\pi}{7+4\sqrt3}\implies R=5\sqrt{\frac3{7+4\sqrt3}}$$
Let now $\;K\;$ be the intersection point between the circle (with center $\;O\;$ ,say) and the radius $\;AM\;$, and form the $\;30-60-90\;$ straight-angle triangle $\;KOM\;$ ,so that
$$R=KO=\color{red}{\frac{\sqrt3}2}\,MO\implies MO=\color{red}{10}\sqrt{\frac1{7+4\sqrt3}}$$
so finally, the radius $\;r\;$ of the circular sector is
$$r=MO+R=\frac{5(\color{red}2+\sqrt3)}{\sqrt{7+4\sqrt3}}=\color{red}5\;\;\left(\text{because}\;\;(2+\sqrt3)^2=7+4\sqrt3\ldots\right)$$
Best Answer
Call $\;O\;$ the center of the big circle from where the circular sector is taken, and let $\;M\;$ be the little circle's center with radius $\;r\;$, and $\;A,B\;$ the two tangent points between circle $\;M\;$ and the the two radii formining the sector.
If we put $\;t:=$the distance between $\;O\;$ and the inner circle, then taking the straight angle triangle $\;\Delta AMO\;$ we get:
$$\sin x=\frac r{r+t}\;\;\text{and also}\;\;r=R-(t+r)\implies t=R-2r\implies$$
$$r=(r+R-2r)\sin x\implies r=(R-r)\sin x\implies r=\frac{R\sin x}{1+\sin x}$$
so finally
$$S=\pi R^2\frac{\sin^2x}{(1+\sin x)^2}$$