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Homework Help: Quick shpere intersection question?

  1. Aug 24, 2009 #1
    quick shpere intersection question??

    1. The problem statement, all variables and given/known data

    Use the given info to answer the following questions

    a. find an equation of the sphere with the give center and radius.

    center (1,-11,3) radius 5

    answer is (x-1)^2 + (y+11)^2 + (z-3)^2 = 25

    where im stuck is the next question

    b. what is the intersection of this sphere with the yz-plane? (they want an equation)

    i know a^2 + b^2 + c^2 = R^2
    but now we are cutting down the plane so im stuck from here

    3. The attempt at a solution

    i figured the yz plane would only have the y and z portion of the equation from part a. so i did that and got (y+11)^2 + (z-3)^2 = 25
    that is wrong. im sure it is a simple formula but could not find it online.
    so any help would be great!!!!


    **smell that?? oh... that is the smell of a new semester. (aka death!!!)**
  2. jcsd
  3. Aug 24, 2009 #2


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    Re: quick shpere intersection question??

    While I am not well versed in 3d spheres and such. On the yz plane, I believe x=0 so you'd have an extra term (-1)2. But your answer would be a circle in y and z since if you look from an above view of the yz plane, you'd see...well a circle.
  4. Aug 24, 2009 #3
    Re: quick shpere intersection question??

    The intersection is circle, but its radius is not equal to 5.

    Since the circle is lying on the yz plane so that the center of the circle is equal to (0,-11,3)

    The equation of the circle will be [itex](y+11)^2+(z-3)^2=r^2[/itex]

    Now the distance between the centers is 1. And the radius of the sphere R=5.

    Using Pythagorean theorem [itex]r^2=5^2-1^2=24[/itex]

    I guess the final equation would be [itex](y+11)^2+(z-3)^2=24[/itex]

  5. Aug 24, 2009 #4
    Re: quick shpere intersection question??

    ok... that makes sense. forgot a circle gets smaller haha
    thanks a lot
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