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  1. #1
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    equation of a circle with...

    Hi,

    Forgot how to do this.


    The line y = -1 is a tangent to a circle which passes through (0, 0) and (6, 0).

    Find the equation of the circle.

    Any help appreciated
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by enji333 View Post
    Hi,

    Forgot how to do this.


    The line y = -1 is a tangent to a circle which passes through (0, 0) and (6, 0).

    Find the equation of the circle.

    Any help appreciated
    Well the circle equation is
    (x - h)^2 + (y - k)^2 = r^2

    So putting the two points in gives
    h^2 + k^2 = r^2
    and
    (6 - h)^2 + k^2 = r^2

    The circle is tangent to the line y = -1. This tells us a few things. First, the circle has a point (X, -1) on it:
    (X - h)^2 + (-1 - k)^2 = r^2

    We also know that the derivative of the circle equation is 0 at y = -1, since y = -1 is a horizontal line. So taking the derivative of the circle equation we get
    2(X - h) + 2(y - k) \frac{dy}{dx} = 0

    The derivative at (X, -1) is 0, so this becomes
    2(X - h) = 0

    Thus you need to solve the system of equations:
    h^2 + k^2 = r^2

    (6 - h)^2 + k^2 = r^2

    (X - h)^2 + (-1 - k)^2 = r^2

    2(X - h) = 0

    -Dan
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  3. #3
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    Hi,

    First of all, sorry for double posting

    About your solution. I'm not sure about the final answer you gave. Would you be able to show me the final answer in the form (x - a)^2 + (y - b)^2 = r^2 ?

    Appreciate help again, thanks for your answer also
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by enji333 View Post
    Hi,

    First of all, sorry for double posting

    About your solution. I'm not sure about the final answer you gave. Would you be able to show me the final answer in the form (x - a)^2 + (y - b)^2 = r^2 ?

    Appreciate help again, thanks for your answer also
    I never gave you a solution. I gave you four equations in the unknowns X, h, k, and r. I'm leaving it for you to solve them. If you have difficulties, please post what you've been able to come up with, and either I or someone else will help you from there.

    -Dan
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  5. #5
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    Quote Originally Posted by enji333 View Post
    The line y = -1 is a tangent to a circle which passes through (0, 0) and (6, 0).

    Find the equation of the circle.
    Draw a rough sketch. You can proof by rules of symmetry that the tangent point must be T(3, -1).

    Use the equation of the circle. Plug in the coordinates of the 3 points and solve the system of equations for h, k, r.

    For confirmation only: (h, k, r) = (3, 4, 5)
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