Results 1 to 8 of 8

Math Help - Two circles with the centre on each others circumference

  1. #1
    Junior Member
    Joined
    Oct 2005
    Posts
    54

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Glaysher's Avatar
    Joined
    Aug 2006
    From
    Newton-le-Willows
    Posts
    224
    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.
    Attached Thumbnails Attached Thumbnails Two circles with the centre on each others circumference-intersection.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2005
    Posts
    54
    I have to think about this one for a while.
    Thanks for the answer, there may be some more questions later.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    648
    Hello, a4swe!

    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<br />

    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}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648

    Lightbulb 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?)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,746
    Thanks
    648
    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.

    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: June 26th 2011, 02:28 PM
  2. Circumference
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 29th 2010, 06:50 PM
  3. Prove that every rigid motion transforms circles into circles
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: February 11th 2010, 06:00 PM
  4. Replies: 2
    Last Post: October 6th 2009, 08:04 AM
  5. Circumference help!
    Posted in the Geometry Forum
    Replies: 8
    Last Post: September 23rd 2008, 01:14 PM

Search Tags


/mathhelpforum @mathhelpforum