Results 1 to 13 of 13

Math Help - Area Of Intersection Between Two Circles

  1. #1
    Newbie
    Joined
    Jan 2011
    From
    NSW, Australia
    Posts
    15

    Area Of Intersection Between Two Circles

    So I've had a problem on my mind for most of today, and although I'm sure the answer is quite simple, I can't seem to figure out how to go about doing this.

    The question is this: Suppose you had two circles (A and B) of equal size with a radius x. If the two circles were arranged such that the circumference of each circle passed through the center of the other, what would be an expression in terms of x for the area where the two circles overlap?

    Diagram of what I'm talking about is below (A and B are the centers of the circles):



    The area I'm looking for being the middle section.

    Hints on how to go about finding it, or a direct solution would be much appreciated :V
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,504
    Thanks
    1401
    Are you expected to use calculus for this problem?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2011
    From
    NSW, Australia
    Posts
    15
    If that's the only way to solve it, then I suppose so, yes.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,504
    Thanks
    1401
    Well to start with, I would rename the radius \displaystyle r, so that you can use \displaystyle (x, y) for points on the Cartesian plane.

    To start with, what could be the equations of the two circles?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2011
    From
    NSW, Australia
    Posts
    15
    Hmm... interesting. I ran with the idea of giving them equations on a Cartesian plane. Just to make sure the method worked, I assumed the circles had a radius of 1 and tried to find the area. I'm not sure if it's the most efficient way to do things, but my method of working it using what I've learnt thus far in class:

    \displaystyle x^2 + y^2 = 1
    \displaystyle (x + 1)^2 + y^2 = 1

    Solving for y^2 in the first then subbing into the second gives:

    \displaystyle (x + 1)^2 + 1 - x^2 = 1

    \displaystyle 2x + 2 = 1

    x = -\frac{1}{2}

    Then back into equation 1:

    (-\frac{1}{2})^2 + y^2 = 1
    \frac{1}{4} + y^2 = 1
    y = \pm\frac{\sqrt{3}}{2}

    Which means the distance between the points of intersection is \sqrt{3}. A triangle then can be constructed using Point A and the two points of intersection, with the side lengths 1, 1 and \sqrt{3}. Using cosine rule to find the angle at the center:

    \theta = \arccos(\frac{1^2 + 1^2 - \sqrt{3}^2}{2\cdot1\cdot1})
    \theta = \frac{2\pi}{3}

    Finally, the angle and radius can be used to calculate the area of the minor segment that makes up half of the area to be found. Doubling the result will give the total area.

    A = 2\cdot\frac{1}{2}r^2(\theta - \sin\theta)
    A = \frac{2\pi}{3} - \sin(\frac{2\pi}{3})

    Giving a final answer of:

    \displaystyle A = \frac{4\pi - 3\sqrt{3}}{6}

    Does the process and working seem correct? Or have I made a mistake somewhere in this? (I'll attempt to do it with an unknown radius r like I originally planned, next.)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,504
    Thanks
    1401
    Your radius is not 1, it's variable. Call it \displaystyle r or something else...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2011
    From
    NSW, Australia
    Posts
    15
    Yeah I know. As I said in the post, I was just doing it under the assumption of 1 to figure out a method. If that working is correct, I'll attempt it using the variable.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2011
    From
    New York
    Posts
    3

    Keep it Simple

    you don't need calculus to solve this problem. let's break it down into simple parts: slice the area you're looking at down the middle (eg connecting the two points where the circles intersect each other). now you have two symmetric sections of a circle. Solve for the area of one of them, and multiply by 2.

    to solve for the area of a segment, solve for the area of the sector, and subtract the area of a triangle.

    first let's solve for the angle of the sector:

    we have the radius = 1, and because we're dealing with a symmetric arrangement, we can make a right triangle with the radius as hypontenuse, and half the radius as the short leg. We can solve for half the angle of the sector (call this theta) by taking cos(theta) = half radius/radius. Note that the radius drops out so we end up with a generalized equation regardless of what the radius is, that lets us solve for the angle. (we will need an actual radius later to solve for the specific area, but we use "r" as has been suggested as the variable for that so any quantity can be plugged in later)

    Therefore cos(theta) = .5;

    arccos(.5) = theta = 60 degrees. two times theta = 120 degrees = the angle of our sector.

    Now compute the area of the sector, and subtract the area of the triangle. We can quickly get the area of the triangle by doubling the area of the smaller triangle described by the radius as hypontenuse w half radius as one side, and then doubling. Thinking about this it becomes obvious that if you take two right triangles and put them together you can create a rectangle of sides radius x half-radius, so the total area of the larger triangle that will be subtracted from the sector is half a radius in square units for area.

    now to compute the area of the sector: 120 degrees is 120/360 = 1/3 of the area of the whole circle. [(pi)*(radius^2)]/3 = the area of the sector.

    [(pi*r^2)/3] - r/2 = half of the area that you are seeking, so multiply by 2:

    2[(pi)r^2]/3 -r = the area of the overlapped circles as shown.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    909
    Thanks
    27
    Hi myfriendzac,
    You need to recalculate the triangle area to be subtracted from the sector area to get the segment area.



    bjh
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jan 2011
    From
    New York
    Posts
    3
    checking my work...
    the altitude of the triangle is half a radius, and the hypotenuse is 1... i took the side of the rectangle to be 1 when that is effectively the diagonal. You're right BJ. The side of the rectangle is the square root of 1-(1/2)^2 = sqrt(1-1/4) = (3/4)^.5 = .866025.

    This makes the area of that triangle to be subtracted from the sector = .5r * .866025r = r(.5*.866025) = .433013r

    Therefore, the corrected area for the overall problem is: 2[(pi)r^2]/3 -.866025r

    BJ, let me know if that corrects things.

    cheers,

    Zac.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by MCXD View Post
    So I've had a problem on my mind for most of today, and although I'm sure the answer is quite simple, I can't seem to figure out how to go about doing this.

    The question is this: Suppose you had two circles (A and B) of equal size with a radius x. If the two circles were arranged such that the circumference of each circle passed through the center of the other, what would be an expression in terms of x for the area where the two circles overlap?

    Diagram of what I'm talking about is below (A and B are the centers of the circles):



    The area I'm looking for being the middle section.

    Hints on how to go about finding it, or a direct solution would be much appreciated :V
    If you join points A and B to the top point of intersection, call it C,
    you will have an equilateral triangle between A, B, C.

    Calculate the area of sector ACB in the left circle with A as centre...

    \frac{1}{2}x^2\frac{\pi}{3}

    Calculate the area of sector BCA, same result.

    Add the areas of both sectors and subtract the area of the equilateral triangle,
    since we have one too many in the area expression.

    x^2\frac{\pi}{3}-\frac{1}{2}x^2sin\frac{\pi}{3}

    Double this result.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    909
    Thanks
    27
    Hi Zac,
    Your corrected area is rad 3/2 r. The correct answer is rad3/2 r^2


    bjh
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Jan 2011
    From
    NSW, Australia
    Posts
    15
    Thanks for the help everyone, and for showing me some much more elegant solutions. Working off Archie Meade's solution I get:

    \displaystyle A = \frac{r^2(4\pi - 3\sqrt{3})}{6}

    Awesome. Thanks again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area of intersection of two circles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 13th 2011, 07:36 PM
  2. intersection of circles
    Posted in the Algebra Forum
    Replies: 9
    Last Post: March 31st 2011, 01:57 PM
  3. Replies: 2
    Last Post: October 6th 2009, 08:04 AM
  4. Intersection of circles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 21st 2009, 07:20 AM
  5. Usable trilateration? (Circles intersection)
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: September 17th 2008, 01:04 PM

Search Tags


/mathhelpforum @mathhelpforum