Area of specific region inside a circle i

Question

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.

Answer 1

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.

Answer 2

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 $d_1$.

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

2. 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 $(a,b)$).

3. 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.

Answer 3

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.