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Thread: Intersection of parametric equations

  1. #1
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    Intersection of parametric equations

    Find the sum of all possible values of the constant such that the graph of the parametric equations




    intersects the graph of the parametric equations



    at only one point.




    I tried using sin^2+cos^2=1 again, ending up with

    x^2 = 4 + 16 cos^2(s) + 16 cos(s)

    y^2 = k^2 + 16 sin^2(s) - 8k sin(s)

    Thus

    x^2 + y^2 = 16(1) + 4 + k^2 + 16 cos(s) - 8k sin(s)

    From here I wondered if I should substitute in sin(s)= (k-y)/4 but would that just take me in circles?

    Thank you
    Last edited by orange; Dec 16th 2013 at 07:29 PM.
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  2. #2
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    Re: Intersection of parametric equations

    These are both equations of circles. Find those equations. Find their centers. The only way that two circles intersect in only one point is if the distance from the center of one circle to the center of the second circle is equal to the length of the sum of their respective radii.
    Thanks from romsek
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  3. #3
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    Re: Intersection of parametric equations

    If you plot these you'll see much quicker what to do.

    Both of these parametric equations represent circles. I leave it to you to find their centers and radii. Adjusting k changes things from having 0 points of intersection, to 1, to 2. The case of 1 point of intersection is where the circles intersect at a tangent point. Once you get your equations into standard form for circles solving for k should be pretty straightforward.

    spoiler
    k = -3 +/- sqrt[6] so their sum is -6
    Last edited by romsek; Dec 16th 2013 at 08:18 PM.
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  4. #4
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    Re: Intersection of parametric equations

    convert both the equations to Cartesian equations nd we get
    (x-1)^2 + (y+3)^2 = 1 Center ( 1, -3 ) radius = 1
    and (x-2)^2 + (y-k)^2 = 16; Center ( 2, k ) radius = 4
    Now you can proceed as explained by slipEternal and romsek
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  5. #5
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    Re: Intersection of parametric equations

    Thank you all so much!

    Aligning the center (2,k) with the point (2,-3) on the smaller circle, I ended up with a right triangle (hypotenuse: 5, connecting each center) and found that the center of the big circle is 2 root(6) away from the point (2,-3) on the smaller one.

    So am I correct in thinking k= 2root(6)-3 ?
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  6. #6
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    Re: Intersection of parametric equations

    There are two k's that solve this. One is a circle below the big one. The other it is above it.

    Redo this and be a bit more careful. Your value of k above isn't quite correct. Then find the other k and add them.

    The answer is under the "spoilers" line in one of my prior posts on this thread. Highlight the white text to see it, but work it out on your own first.
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