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Math Help - Circle-Cirlce tangent in three dimensions

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    Circle-Cirlce tangent in three dimensions

    My problem is to find two circles in 3D space that share a tangent vector. The only thing known is a a point and tangent on each circle and the radius for both circles.

    The problem can be visualized with this image/video:



    Known:
    P_a - Point on Circle A
    \vec{B_a}- Circle tangent at point Pa, normalized
    R_a - Radius of Circle A
    P_b - Point on Circle B
    \vec{B_b}- Circle tangent at point Pb, normalized
    R_b - Radius of Circle B

    Wanted:
    D_a - Circle-Circle tangent point on Circle A
    D_b - Circle-Circle tangent point on Circle B

    We know the following:
    (C_a-D_a)\cdot(D_b-D_a) = 0
    (C_b-D_b)\cdot(D_b-D_a) = 0
    (C_a-P_a)\cdot\vec{B_a} = 0
    (C_b-P_b)\cdot\vec{B_b} = 0
    |C_a-P_a|=|C_a-D_a|=R_a
    |C_b-P_b|=|C_b-D_b|=R_b
    (C_a-D_a)\cdot(C_b-D_b) = R_a*R_b*\cos{\alpha}
    \alpha - is the angle(unknown) between the Circle A and Circle B plane

    I have tried to solve this by substitution but it gets messy. I'm happy for any suggestion in how to find an analytic solution. Maybe I have missed some geometric relations that could help to find a solution?

    Here is some relations that could be used.

    There is a relation between the given circle tangents and the wanted circle-circle tangent which gives:
    Circle A:
    (D_a-P_a)\cdot(D_a-D_b) = |D_a-P_a|*|D_a-D_b|*\cos{\beta}
    (D_a-P_a)\cdot{B_a} = |D_a-P_a|*\cos{\beta}
    This gives:
    |D_a-D_b| = \frac{(D_a-P_a)\cdot(D_a-D_b)}{(D_a-P_a)\cdot{B_a}}
    Doing same thing for Circle B gives:
    |D_a-D_b| = \frac{(D_b-P_b)\cdot(D_a-D_b)}{(D_b-P_b)\cdot{B_b}}
    Combining this gives:
    (D_a-P_a)\cdot(D_a-D_b)*(D_b-P_b)\cdot{B_b} = (D_b-P_b)\cdot(D_a-D_b)*(D_a-P_a)\cdot{B_a}
    Last edited by obidobi; August 19th 2010 at 07:04 PM.
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