# Two circles with the centre on each others circumference

• Sep 1st 2006, 12:58 AM
a4swe
Two circles with the centre on each others circumference
Problem: Two circles, both with the radius of R (and in the same plane) intersect so that the centre of one circle lies on the circumference of the other circle. Calculate the area inside BOTH of the circles.
• Sep 1st 2006, 01:59 AM
Glaysher
If this is the case and the centre of one circle (circle 1) lies on the circumference of the other (circle 2) then since the radi are the same the centre of circle 2 lies on the circumference of circle 1

Area of circle 1 = $\pi r^2$ = Area of circle 2

Two circles considered seperately = $2\pi r^2$

The trick is to find the area of the intersection

The part where the intersection is looks like the image attached

By finding the angle in the diagram I can find the area of the secto between the two radi of circle 1

By drawing a vertical line downwards from the top point down to the horizontal line I create two right angled triangles with hypotenuse $r$ and adjacent (to half the angle I want) $\frac{r}{2}$.

Call the angle I want $\theta$

Then $\cos \frac{\theta}{2} = \frac{\frac{r}{2}}{r}$ $= \frac{1}{2}$

$\frac{\theta}{2}= \frac{\pi}{3}$ in radians

$\theta = \frac{2\pi}{3}$

Area Of Sector = $\frac{1}{2}r^2 \frac{2\pi}{3}$

Now you can do the rest

Find the area of the segment by taking away the area of the triangle formed by the radi

The area of the segment is half the area of the intersection so multiply by two and take away from $2\pi r^2$

EDIT: You will probably need to draw more diagrams to follow what I've done. It's not easy to draw them all so that I can post them here with the software I have.
• Sep 1st 2006, 04:44 AM
a4swe
Thanks for the answer, there may be some more questions later. :)
• Sep 1st 2006, 06:27 AM
Soroban
Hello, a4swe!

Quote:

Two circles, both with the radius of R (and in the same plane) intersect
so that the centre of one circle lies on the circumference of the other circle.
Calculate the area inside both of the circles.

The intersection is a lens-shaped region.
Code:

            *         * /:::*       *  /::::::*         /:::::::::     *  /::::::::::*       /::::::::::::       /::::::::::::::     */::::::::::::::*     * 120°::::::::::*     *\::::::::::::::*       \::::::::::::::       \::::::::::::     *  \::::::::::*         \:::::::::       *  \::::::*         *-\:::*             *

In Glaysher's excellent diagram, we see two equilateral triangles.
. . Hence, we have a 120° sector plus two 60° segments.

Since the sector occupies one-third of the circle: . $A_{\text{sector}} \:=\:\frac{1}{3}\pi R^2$

The area of a segment is: . $\text{(Area of 60}^o\text{ sector)} - \text{(Area of triangle)}$
. . . $A_{\text{segment}}\;= \;\frac{1}{6}\pi R^2 - \frac{\sqrt{3}}{4}R^2 \;= \;\frac{2\pi - 3\sqrt{3}}{12}R^2
$

Hence, the area of the two segments is: . $A_{\text{segments}} \:=\:\frac{2\pi-3\sqrt{3}}{6}R^2$

Therefore, the area of the intersection is:

. . $A \;= \;\frac{1}{3}\pi R^2 + \frac{2\pi-3\sqrt{3}}{6}R^2 \;= \;\boxed{\frac{4\pi - 3\sqrt{3}}{6}R^2}$

• Sep 4th 2006, 03:34 AM
malaygoel
A somewhat similar question
Hello to all of you!
I have a question relating to circles. It is like this:
There is a grass field in the circular shape, with the fence along the boundary. A goat is tied to the fence(at a stationary point) with a rope such that she can graze half of the area of the field. The problem is to find the ratio of the length of the rope to the radius of the circular field.
(Can it be done without using calculus?)
• Sep 4th 2006, 04:46 AM
CaptainBlack
Quote:

Originally Posted by malaygoel
Hello to all of you!
I have a question relating to circles. It is like this:
There is a grass field in the circular shape, with the fence along the boundary. A goat is tied to the fence(at a stationary point) with a rope such that she can graze half of the area of the field. The problem is to find the ratio of the length of the rope to the radius of the circular field.
(Can it be done without using calculus?)

This has to be done numerically, you end up with a mixed
algebraic/transcendental equation to solve, which I believe has no
known elementary solution.

It can be solved numerically to whatever precision you want using
the binary chop or search algorithm, which does not require any
calculus.

RonL
• Sep 4th 2006, 05:27 AM
Soroban
Hello, Malay!

Welcome back!

As the Cap'n pointed out, it is a classic problem: "The Half-Pastured Goat".

Even with Calculus, it cannot be solved by elementary methods.
As RonL said, the solution can be approximated by numerical means.

• Sep 4th 2006, 05:43 AM
CaptainBlack
Quote:

Originally Posted by Soroban
Hello, Malay!

Welcome back!

As the Cap'n pointed out, it is a classic problem: "The Half-Pastured Goat".

Even with Calculus, it cannot be solved by elementary methods.
As RonL said, the solution can be approximated by numerical means.

Not only is it a classic problem it has more variants than you can
shake a stick at.

See here for a discussion of the problem and follow the links from there (and the links from the links ...) for variants.

RonL