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Math Help - Spheres

  1. #1
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    Spheres

    I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time

    The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:
    Sphere S: x^2+y^2+z^2-2y-22z-103=0
    Point C, centre of the circle of intersection: 6/7/8
    and the plane P 2x+2y-z-18=0

    My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)
    Then I invented the line l, which goes from C farther on the plane an eventually hits the spehere.
    l= (6,7,8)+t(1,-1,0)
    To find where it hits the plane, I inserted the line in the equation of the sphere
    (1+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0
    As result, I get t^2=(70)^1/2
    This value inserted for t and than substracted the beginning point C, the radius of my circle gets (140)^1/2 which is the wrong solution. The result must be 12 (so (144)^1/2).
    As many times as I perform the calculation, I get the wrong solution, where is the flaw of my approach??
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  2. #2
    Member mohammadfawaz's Avatar
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    Hello Schdero,

    Try to find any point on the intersection between the plane and the sphere. For example, take x = 0 and replace in the equation of the plane and in the equation of the sphere. You get a system of 2 equations with two unknowns. Solve for y and z. If you did not succeed, try to take y = 0 and repeat. Now the radius is simply the distance between C and that point of intersection.

    Hope this helps
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  3. #3
    Super Member Failure's Avatar
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    Quote Originally Posted by Schdero View Post
    I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time

    The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:
    Sphere S: x^2+y^2+z^2-2y-22z-103=0
    Point C, centre of the circle of intersection: 6/7/8
    and the plane P 2x+2y-z-18=0

    My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)
    No, I think you are headed in the wrong direction: the problem is much simpler than you think. First, determine the center and radius of the sphere by completing the square. I get S:\; (x-0)^2+(y-1)^2+(z-11)^2=15^2. So the center of the sphere is  M(0/1/11) and its radius is r=15.
    Since the center C(6/7/9) of the circle of intersection of sphere and plane is given, the radius \rho of the circle of intersection is one leg of a right triangle with hypotenuse r=15 and the other leg |MC|. Thus, \rho=\sqrt{r^2-|MC|^2}=\sqrt{15^2-81}=12.
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  4. #4
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    Thank you, your solution is much simpler!

    Schöne obe no ;-) (hope your not only swiss, but swiss-german^^)
    Last edited by Schdero; April 3rd 2010 at 01:41 PM.
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  5. #5
    Super Member Failure's Avatar
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    Quote Originally Posted by Schdero View Post
    I'm very sory for asking another question, but I'm just preparing a final exam, which means that i solve something like 20 Problems a day, so I get confused from time to time

    The Sphere S intersects the plane P in a circle. Find the radius of the circle. Given are:
    Sphere S: x^2+y^2+z^2-2y-22z-103=0
    Point C, centre of the circle of intersection: 6/7/8
    and the plane P 2x+2y-z-18=0

    My idea was first to find one direction vector of the plane, so for instance (1,-1,0) (one vector which lies on the plane)
    Then I invented the line l, which goes from C farther on the plane an eventually hits the spehere.
    l= (6,7,8)+t(1,-1,0)
    To find where it hits the plane, I inserted the line in the equation of the sphere
    ({\color{red}1}+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0
    As result, I get t^2=(70)^1/2
    The above equation is wrong, it should be
    ({\color{red}6}+t)^2+(7-t)^2 +8^2-2(7-t)-22(8)-103=0
    With this modification you get that t_{1,2}=\pm \sqrt{72}=\pm 6\sqrt{2}

    This value inserted for t and than substracted the beginning point C, the radius of my circle gets (140)^1/2 which is the wrong solution. The result must be 12 (so (144)^1/2).
    As many times as I perform the calculation, I get the wrong solution, where is the flaw of my approach??
    Your approach is ok (although too complicated for my taste): it is only your calculation that was wrong.
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  6. #6
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    I found that out too. Im sorry about that, i calculated it four times and got (70)^1/2=t every single time.
    Thanks anyway
    schdero
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