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Math Help - Finding the radius of a Circle Touching a Parabola and the Axis

  1. #1
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    Thumbs up Finding the radius of a Circle Touching a Parabola and the Axis

    Hi Forum!

    I have a little problem with a question:

    What is the radius of a a circle that touches the parabola 1/2x at (2,2) and the x-axis?

    Here the derivative of our parabola is y=x
    And if we get
    (x-2)^2+(y-2)^2=(x-v)^2+y^2

    This could give us the common point between those two circles; that would be the center of the circle that we want.
    But this gives me a strange equation, that I really don't know what it is for.

    Or,
    Get the center (a,b)
    and use it at the circle equation.
    (x-a)^2+(y-b)^2=r^2

    But where does (2,2) enters here?
    I saw something about using both at the same equation, like:

    (a-2)^2+(b-2)^2=r^2
    Is this possible? Aren't x and y fixed values here?

    Thanks! I can't find a good source to research on this.
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  2. #2
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    Quote Originally Posted by Zellator View Post
    What is the radius of a a circle that touches the parabola 1/2x at (2,2) and the x-axis?

    Here the derivative of our parabola is 2
    And if we get
    (x-2)^2+(y-2)^2=(x-v)^2+y^2 What is v?

    This could give us the common point between those two circles; that would be the center of the circle that we want. Which two circles?
    But this gives me a strange equation, that I really don't know what it is for. Why is it strange?
    It's not easy to understand your thought.

    Hope this helps.

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  3. #3
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    Oh, sorry to not make myself clear, emakarov.

    v is the x coordinate of the place where the circle touches the x-axis. The Xo in your graph.
    I read somewhere that you could get get center of the circle by using two circles in the places where we have points, the parabola and the x-axis.
    Using this formula I'll get a equation with some unwanted X's and Y's and V's, so it's probably not the best idea to use it.

    I really thought about using a perpendicular here. But what is the easiest way to get it?

    I would use it like f(x-2)+2 using the translation principles.
    y'=-(1/2)x -2

    So

    y'=-(1/2)(x-2)+2=-(x/2)+3


    But what about the tangent line?

    To get it you used y-2=2(x-2) right?
    I was kind of confused in that terms too, since I hadn't seen calculus in some time.

    Ok, I will give it a go one more time.
    Thanks again, emakarov!
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  4. #4
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    You have two equations on x0, y0: the first says that the distance from (x0,y0) to (2,2) equals the distance from (x0,y0) to (x0,0), and the second says that (x0,y0) lies on the graph of y = 3 - x/2.
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  5. #5
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    Quote Originally Posted by emakarov View Post
    You have two equations on x0, y0: the first says that the distance from (x0,y0) to (2,2) equals the distance from (x0,y0) to (x0,0), and the second says that (x0,y0) lies on the graph of y = 3 - x/2.
    Hi emakarov!
    I've managed to do it!!!
    Yes, it was quite easy with your explanation! Thank you so much!
    There's another way that I don't understand well


    What's that -1 and that 2?
    I'm not sure, but the b^2-5b+5 could be acquired in some way, replacing a for b in the first equation.
    This way it was answered in terms of y, it seems easier to use x though!

    Thanks emakarov!
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