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Math Help - Intersection of Two Spheres

  1. #1
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    Intersection of Two Spheres

    For quite sometime I have been intrigued by the curve created by the intersection of two spheres with the following properties/conditions:

    Let both spheres have the same radius. Let's say a radius of 2.
    Let one sphere be centered at (0,0,0) and the other sphere be centered at
    (2,2,2).

    I think the equations for the two spheres are:
    x^2+y^2+z^2=4
    and
    (x-2)^2+(y-2)^2+(z-2)^2=4

    I have seen a method finding the intersection when both spheres have the same coordinates on the y and z axis and differing coordinates on the x axis. This involves combining the two equations and solving for x.
    This would yield a circle parallel to the yz plane.

    So, here is what I did:
    (x-2)^2+(y-2)^2+(z-2)^2=x^2+y^2+z^2
    Multiplying through and rearranging I got
    x+y+z=3.
    I think this is the equation of the plane where the two spheres intersect.

    I am not sure what to do next.
    I have read that it will be a circle and that there will be two equations describing the curve.(Now that i think about it i guess the intersection of 2 spheres will always be a circle)
    I had built a model about 10 years ago out of balsa wood and the curve really did not look circular. But maybe that was due to the inaccuracies of the model.
    How do I find an equation that describes this curve.
    I looked in my old calculus book but could find nothing.

    Any help would be greatly appreciated.
    Last edited by kid funky fried; January 3rd 2010 at 08:16 PM. Reason: semicircle to circle, add exponent
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  2. #2
    MHF Contributor matheagle's Avatar
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    You have x+y+z=3, but you also have the original constraints.

    IF you shift the second center away from (2,2,2) this will be easier to observe.

    Look at (x-2)^2+y^2+z^2=4 and x^2+y^2+z^2=4

    Then, when you solve for the intersection you have x=1, but NOT all points on the plane x=1.

    Letting x=1 into either/both of your orginal constraints you have y^2+z^2=3

    so the intersection of these two spheres is x=1 and y^2+z^2=3

    which is a circle in the plane x=1.

    I was playing around with the algebra in your case and it seems to be a circle.

    Insert z=3-x-y into the sphere x^2+y^2+z^2=4 and see what you get.
    Last edited by matheagle; January 3rd 2010 at 08:28 PM.
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    Member Abu-Khalil's Avatar
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    Quote Originally Posted by matheagle View Post
    I was playing around with the algebra in your case and I'm not confident that you have a circle.
    From where i see it, it's the same than having one sphere centered at \left(2\sqrt{2},0,0\right). Just need a rotation.
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  4. #4
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    Quote Originally Posted by kid funky fried View Post
    For quite sometime I have been intrigued by the curve created by the intersection of two spheres with the following properties/conditions:

    Let both spheres have the same radius. Let's say a radius of 2.
    Let one sphere be centered at (0,0,0) and the other sphere be centered at
    (2,2,2).
    The intersection two spheres, or of any plane with a sphere, is either empty or a circle. In this case, your two spheres each have radius 2 and the distance between their centers is \sqrt{4+ 4+ 4}= 2\sqrt{3}< 4 so they intersect in a circle. That circle will line in the plane that perpedicularly bisects the line between their centers. The midpoint of that line is (1, 1, 1) and a vector in the direction of that line is \vec{i}+ \vec{j}+ \vec{k}. The equation of that plane is 1(x-1)+ 1(y- 1)+ 1(z- 1)= 0 or x+ y+ z= 3. Since the distance from (2, 2, 2) to (1, 1, 1) is \sqrt{3} and the radius of each sphere is 2, the radius of the circle of intersection is one leg of a right triangle having one leg of length \sqrt{3} and hypotenuse 2: the radius is \sqrt{(\sqrt{2^2- (\sqrt{3})^2}= 1. That is, the intersection is a circle with center at (1, 1, 1), radius 1, in the x+ y+ z= 3 plane.


    I think the equations for the two spheres are:
    x^2+y^2+z^2=4
    and
    (x-2)^2+(y-2)^2+(z-2)^2=4

    I have seen a method finding the intersection when both spheres have the same coordinates on the y and z axis and differing coordinates on the x axis. This involves combining the two equations and solving for x.
    This would yield a circle parallel to the yz plane.

    So, here is what I did:
    (x-2)^2+(y-2)^2+(z-2)^2=x^2+y^2+z^2
    Multiplying through and rearranging I got
    x+y+z=3.
    I think this is the equation of the plane where the two spheres intersect.

    I am not sure what to due next.
    I have read that it will be a circle and that there will be two equations describing the curve.(Now that i think about it i guess the intersection of 2 spheres will always be a circle)
    I had built a model about 10 years ago out of balsa wood and the curve really did not look circular. But maybe that was due to the inaccuracies of the model.
    How do I find an equation that describes this curve.
    I looked in my old calculus book but could find nothing.

    Any help would be greatly appreciated.
    Last edited by HallsofIvy; January 3rd 2010 at 01:36 AM.
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    Thanks

    Thanks to all. I appreciate the timely response.
    A follow up question-

    For a given curve in space is there a way to find an equation or set of equations to define the curve?
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  6. #6
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    I've been working on this for some time- mostly going down a blind alley!

    As I said before, the intersection of two spheres, each of radius 2, with centers at (0, 0, 0) and (2, 2, 2), is a circle of radius 1, with center at (1, 1,1), in the plane x+ y+ z= 3.

    As for finding parametric equations for that circle, matheagles remark, "IF you shift the second center away from (2,2,2) this will be easier to observe", gives a good idea. (Abu-Kahlil said much the same thing [his " 2\sqrt{2}" should have been " 2\sqrt{3}"] but it was his "Just need a rotation" that sent me down a blind alley looking for rotation matrices!) If the two spheres were centered at (0, 0, 0) and (2\sqrt{3}, 0, 0), which is the same distance from (0, 0, 0), then their intersection is a circle of radius 1 with center at (\sqrt{3}, 0, 0), in the plane x= \sqrt{3}, perpendicular to the x-axis. The parametric equations for that circle are easy- x= \sqrt{3}, and we can use the "usual" parametric equations for a circle of radius 1, y= cos(t) and z= sin(t).

    Now, we need to find a linear transformation that will change one geometric situation into the other (it doesn't have to be linear but that is simplest). That is, a linear transformation that will take the vectors <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> into axes appropriate to this problem.

    The line through the center of one circle, perpendicular to its plane, is the x-axis, y= z= 0, while the line through the center of the other, perpendicular to its plane is x= y= z. A vector in that direction is <1, 1, 1> and a unit vector in that direction is \left<1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}\right>.

    For a second coordinate axis, we need vector perpendicular to that, so in the plane x+ y+ z= 0. Just about any will do, so take z= 0. That makes the equation x+ y= 0 or y= -x so <-1, 1, 0> will do and a unit vector in that direction is \left<-1/\sqrt{2}, 1/\sqrt{2}, 0\right>.

    The third axis must be perpendicular to both, so we take the cross product of the first two: \left<1/\sqrt{6}, 1/\sqrt{6}, -2/\sqrt{6}\right>.

    Now we need a matrix that will map <1, 0, 0> into \left<1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3}\right>, <0, 1, 0> into \left<-1/\sqrt{2}, 1/\sqrt{2}, 0\right>, and <0, 0, 1> into \left<1/\sqrt{6}, 1/\sqrt{6}, -2/\sqrt{6}\right>.

    It is well known that the matrix that maps three basis vectors into three given vectors is the matrix having those given vectors as columns:
    \begin{bmatrix} \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\end{bmatrix}.

    Take the product of that matrix with each of \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}, and \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} to see why that works.

    To change the parametric equations x= \sqrt{3}, y= cos(t), z= sin(t) to this new "basis" and so give equations for the original circle, we multiply
    \begin{bmatrix} \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\end{bmatrix}\begin{bmatrix}\sqr  t{3} \\ cos(t) \\ sin(t)\end{bmatrix} = \begin{bmatrix} 1-\frac{1}{\sqrt{2}}cos(t)+ \frac{1}{\sqrt{6}}sin(t) \\ 1+ \frac{1}{\sqrt{2}}cos(t)+ \frac{1}{\sqrt{6}}sin(t) \\ 1- \frac{2}{\sqrt{6}}sin(t)\end{bmatrix}.

    That is, parametric equations for the circle forming the intersection of the two spheres are x= 1- 1/\sqrt{2} cos(t)+ 1/\sqrt{6}sin(t), y= 1+ 1/\sqrt{2} cos(t)+ 1/\sqrt{6} sin(t), and z= 1- 2/\sqrt{6} sin(t).

    It is easy to see that those do give the correct figure.
    (x-1)^2= 1/2 cos^2(t)- 1/\sqrt{3} sin(t)cos(t)+ 1/6 sin^2(t)
    (y-1)^2= 1/2 cos^2(t)+ 1/\sqrt{3} sin(t)cos(t)+ 1/6 cos^2(t)
    (z-1)^2= 2/3 sin^2(t)

    1/2 cos^2(t)+ 1/2 cos^2(t)= cos^2(t), -1/\sqrt{3} sin(t)cos(t)+ 1/\sqrt{3} sin(t)cos(t)= 0, and 1/6 sin^2(t)+ 1/6 sin^2(t)+ 2/3 sin^2(t)= sin^2(t) so
    (x-1)^2+ (y-1)^2+ (z-1)^2= cos^2(t)+ sin^2(t)= 1

    Also, since -\sqrt{2} cos(t)+ 1/\sqrt{2}cos(t)= 0 and 1/\sqrt{6}sin(t)+ 1/\sqrt{6}sin(t)- 2/\sqrt{6}sin(t)= 0, we have x+ y+ z= 1+ 1+ 1= 3.

    The points given by the parametric equations lie on the intersection of that sphere of radius 1 and that plane. Since the center of the sphere, (1, 1, 1) lies on the plane x+ y+ z= 3, there intersection is the circle of radius 1, with center at (1, 1, 1), in the plane x+ y+ z= 3.
    Last edited by HallsofIvy; January 5th 2010 at 04:43 AM.
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  7. #7
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    Thank You

    HallsofIvy,

    Thank you so much. I sincerely appreciate the amount of time that you have spent on this problem.
    I will print your response, take it to work with a couple of my old text books and study until I understand fully.

    Best regards,
    Kid
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