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Thread: Determining the intercept of a circle

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    Determining the intercept of a circle

    Working in the first quadrant only.

    I have a circle defined by the standard and known equation of
    y = (r^2 - x^2)^-0.5

    Where r is the constant radius of the circle.

    y' = -x / (r^2 - x^2)^-0.5

    0 < y < r
    0 < x < r

    I have a tangent line to the circle. I know where it intercepts the X axis at a value greater than d. It is the known variable.

    I need to find a way to determine the slope of the tangent line or the (x,y) coordinate of where the line intercepts the circle. Actually solving one is solving the other.

    intercept = function of distance
    distance = radius plus extra

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    search derivative tangent slope.

    https://gato-docs.its.txstate.edu/sl...t-Line/Finding

    https://www.wyzant.com/resources/ans..._by_c_m_8_2_10

    equation of a line y = mx +b were m is the slope dy/dx.

    Derivative gives the slope at the point on the curve, dy/dx @ [x,y].

    [x,y] is the point on the tangent line, m the slope, solve for b

    b = y - mx
    Last edited by steve_bank; 04-02-2019 at 12:41 AM.

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    Thank you. I did find a solution.

    Instead of using "y = mx + b" I used "y = m(x - c)" where c is the x intercept.

    "y' = -x / (r^2 - x^2)^-0.5" means "m = -x / (r^2 - x^2)^-0.5"

    Therefore by substitution I got

    y = (-x / (r^2 - x^2)^-0.5)(x - c)

    Setting that equal to the original equation for a circle I got

    (r^2 - x^2)^0.5 = (-x / (r^2 - x^2)^0.5)(x - c)

    Eventually by basic algebra I simplified that to

    r^2 = x*c
    x = r^2 / c

    c is the intercept, so it is r + a, so we get

    x = r^2 / (r + a)

    Since r is a constant and a is the variable, I finally defined x in terms of a.

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    ok.

    Plot a circle and the tangent line on top of it from your equations. You can probably do it in excel. See how it looks.

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    y^2 = [r^2 - x^2]^1/2

    Applying the chain rule for u^n Du^n = n* [u]^n-1 * du

    m = -x/sqrt(r^2-x^2)

    At x = 0 m = 0 and you get a horizontal line tangent at y = r.

    for r = 1 the midpoint in the 1st quadrant the point is (.707,.707).

    m = -1 y = -x + 1.414

    the x y intercepts are 1.414 and 1.414

    m = -x/sqrt(r^2-x^2)
    At x = r there is a zero. In the limit x goes to infinity the slope goes to infinity giving a vertical line tangent at x = r.

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