Area of specific region inside a circle i Let S be the circle with centre $(0,0)$, radius $r$ units. The chord $C$ of the circle S subtends an angle of 2π/3 at its center. If R represents the region consisting of all points inside S which are closer to C than to circumference of S, then
(1) What is area of region $R$,  and,
(2) In what ratio does $C$ divide the region $R$ ?
I found the equation of a circle intersecting circle radius $r$ for the chord subtending angle $2\pi/3$ at center we should have
$$ x=\dfrac{r}{2} $$
but don't know what to do next. Please help.
 A: Using the symmetry, we can consider a upper semicircle
where $DE$ is a half of the chord.

The chord splits the area in question
in two parts.
Let the points $V_{1,2}$ are the points on the boundary,
for which
the closest points on a circumference are $U_{1,2}$,
and the closest points on a chord are $W_{1,2}$,
and the value of the closed distances are $x_{1,2}$.
Then we can find $x_{1,2}$ in terms of the angles $\phi_{1,2}$ as
\begin{align} 
x_1&=
\tfrac12\,\frac{R(2\cos\phi_1-1)}{\cos\phi_1-1}\tag{1}\label{1}
,\\
x_2&=
\tfrac12\,\frac{R(2\cos\phi_2-1)}{\cos\phi_2+1}\tag{2}\label{2}
.
\end{align}
The coordinates of the point
on the boundary of the first region in therms of $\phi$ are:
\begin{align}
\frac{\tfrac12\,R}{1-\cos\phi}
\cdot(\cos\phi,\, \sin\phi)\tag{3}\label{3}
.
\end{align}
Similarly, the coordinates of the point
on the boundary of the second region in therms of $\phi$ are:
\begin{align}
\frac{\tfrac32\,R}{\cos\phi+1}
\cdot(\cos\phi,\, \sin\phi)\tag{4}\label{4}
.
\end{align}
Now we can find the functions of the boundary curves
\begin{align}
y_1(x)&=
\tfrac12\,\sqrt{R^2+4R\,x}\tag{5}\label{5}
,\\
y_2(x)&=
\tfrac12\,\sqrt{9R^2-12R\,x}\tag{6}\label{6}
.
\end{align}
So the areas can be found as
\begin{align}
S_1&=\int_{-\tfrac14\,R}^{\tfrac12\,R}
y_1(x)\, dx 
=\tfrac{\sqrt3}4\,R^2
\tag{7}\label{7}
,\\
S_2&=\int_{\tfrac12\,R}^{\tfrac34\,R}
y_2(x)\, dx 
=\tfrac{\sqrt3}{12}\,R^2
\tag{8}\label{8}
.
\end{align}
The total area of the region
\begin{align}
S_{\textsf{tot}}
&=
2S_1+2S_2=
\tfrac23\,\sqrt3\,R^2
\tag{9}\label{9}
,
\end{align}
and the chord splits the area as
\begin{align}
S_1:S_2&=3:1
\tag{10}\label{10}
.
\end{align}
Edit
As @Narasimham pointed out,
both boundary lines are parabolas,
with the focus at the origin
and lines
$x=-\tfrac12\,R$
and
$x=\tfrac32\,R$
as directrices.
A: The circle of interest is $x^2+y^2=r^2$. Since there are infinitely many chords which subtend an angle $\frac{2\pi}{3}$ at the center, we might as well assume the chord to be horizontal (slope $0$). By some simple trigonometry , you’ll find that that equation of the chord is $$y=\frac r2$$ Now, let $(a,b)$ with $a^2+b^2\lt 1$ be any point which is closer to $C$ than to the circumference. First, you need to determine the distance of $(a,b)$ from the circumference, which I’ll call $d_1$.

*

*Find the equation of the line joining the center of the circle and $(a,b)$.


*Solve for the point of intersection of this line with the circle (you’ll get two points, of which you must choose the closer one to $(a,b)$).


*Then, apply the formula for the distance between two points to get $d_1$.
Now, $d_2$ (distance of $(a,b)$ from $C$) is just $$|b-\frac r2|$$ Then to get the locus of all such points, we need $$d_2\lt d_1 \implies d_2^2\lt d_1^2$$ You’ll get a quadratic inequality in $b$ from here, which you have to solve by finding its roots. After that, you’ll get that the value of $b$ must be between two functions, something like $$f(a)\lt b\lt g(a)$$ and this will be the required region $R$. Finding the area is simple from here by integration.
This is a lot of work, but it will work. Hope I was able to help.
A: 
When you draw the circle and the chord, it should become simpler to understand. This is a bit difficult to visualize otherwise.
You could pick any chord that subtends an angle of $\frac{2\pi}{3}$ at the center.
For simplicity, let's pick chord (C) with the equation $y = \frac{R}{2}$ where R is the radius of the circle. As you can see, it will subtend an angle of $120^0$ at the center.
Let's pick any point A $(r,\theta)$ inside the circle of radius R.
Now as per problem statement, distance from point A to the line C has to be smaller or equal to the distance from point A to the circumference of the circle.
d = distance from point $A (r, \theta)$ to the line C = $|\frac{R}{2}-rsin\theta|$
r' = distance from point A to the circumference = R-r
As $d \le r', \space |\frac{R}{2}-rsin\theta| \le R-r$.
A few tests for right half of the circle ($x \ge 0$). We can double the area once we find for the right half.
For $r = R, \theta = \frac{\pi}{6}$
For $\theta = -\frac{\pi}{2}, r \le \frac{R}{4}$
For $\theta = \frac{\pi}{2}, r \le \frac{3R}{4}$
You can find out that there is discontinuity at $\frac{\pi}{6}$ and two different curves between
$-\frac{\pi}{2} \le \theta \le \frac{\pi}{6}$ with Area $A_1$ (say) and $\frac{\pi}{6} \le \theta \le \frac{\pi}{2}$ with Area $A_2$ (say).
$\begin{align*}
A_1 &= \int_{-\pi/2}^{\pi/6}\int_{0}^{\frac{R}{2(1-sin\theta)}}r dr d\theta\\\\
A_2 &= \int_{\pi/6}^{\pi/2}\int_{0}^{\frac{3R}{2(1+sin\theta)}}r dr d\theta
\end{align*}$
I hope you can take it from here.
Area of region R = 2($A_1+A_2)$
Ratio in which chord C divides the region R = $\frac{A_1}{A_2}$. It is basically the ratio of area of region R below the chord and above it.
