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Math Help - Finding the center of a cirlcle tangent to a line

  1. #1
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    Arrow Finding the center of a cirlcle tangent to a line

    We have a line with the equation y=(-4/3)x+4.

    The circle has a radius 1, and the top of the circle also has the tangent line y=2, and the bottom has a tangent line y=0.


    We need to find k in the diagram below or at least the point where y=(-4/3)x+4 intersects the circle.

    We are at a loss on how to solve this, but are sure that we're overthinking this. Can anyone help us get started?

    EDIT: the center of the circle should be (k, 1), not (k, 2).

    Last edited by leverin4; March 25th 2009 at 07:33 PM.
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by leverin4 View Post
    We have a line with the equation y=(-4/3)x+4.

    The circle has a radius 1, and the top of the circle also has the tangent line y=2, and the bottom has a tangent line y=0.


    We need to find k in the diagram below or at least the point where y=(-4/3)x+4 intersects the circle.

    We are at a loss on how to solve this, but are sure that we're overthinking this. Can anyone help us get started?


    Is the picture one that you drew or was it given to you. I'm only asking, because the center of a circle usually has coordinates (h,k) and based on the picture, there is really no way that the y coordinate of the center of the circle could be 2, since 2 is the upper tangent. It would make sense that the x coordinate would be 2. Can you verify that?
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  3. #3
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    Quote Originally Posted by mollymcf2009 View Post
    Is the picture one that you drew or was it given to you. I'm only asking, because the center of a circle usually has coordinates (h,k) and based on the picture, there is really no way that the y coordinate of the center of the circle could be 2, since 2 is the upper tangent. It would make sense that the x coordinate would be 2. Can you verify that?
    I drew the circle in a rush, and you're right. The center of the circle should be (k, 1); that was just a typo. My friend and I have kept working while we are waiting, and we think we figured out that k should be equal to 5/2. Can anyone confirm?
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  4. #4
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    possible solution?

    Just to verify--is the y-coordinate of your circle at 2 or at 1?

    The equation for the circle is
    (x-k)^2 + (y-1)^2 = 1.

    The equation for the line is
    y = (-4/3)x + 4.

    Substitute this expression for y into the equation for the circle to get a new equation in terms of only x and k. Expand the terms to get a quadratic in x. Use the quadratic formula to solve for x in terms of k.

    Depending on the value of k, your circle will either intersect the line at exactly two points, exactly one point, or at no points. Try to determine which values of k will result in only one intersection.

    Hope this helps.
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  5. #5
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    You should get two values for k, k = 1,\; k = 7/2. The first is when the line is ontop of the circle, the second below (as in your picture). You could also use calculus having the derivative of the circle equal the slope of the line at some point (x_0,y_0)
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