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Math Help - Circle and tangents

  1. #1
    Super Member dhiab's Avatar
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    Circle and tangents

    ( C ) the circle for equation :
    the straight line for equation : .
    ( m is the real parametre )
    Find m in case : tangent for ( C ) .
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  2. #2
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    Quote Originally Posted by dhiab View Post
    ( C ) the circle for equation :
    the straight line for equation : .
    ( m is the real parametre )
    Find m in case : tangent for ( C ) .
    1. All lines with the equation y = m(x-4) pass through the point P(4,0).

    2. Calculate the coordinates of the points of intersection between all lines and the circle. If the line is a tangent then there exists only one point of intersection, that means the discriminante must be zero:

    Plug in the linear term for y into the 1st equation:

    x^2-2x+(m(x-4))^2-4(m(x-4)) = 0

    x^2(m^2 + 1) - x(8m^2 + 4m + 2) + 16m^2 + 16m = 0

    Solve for x:

    x = - \dfrac{\sqrt{- 4m^2 - 12m + 1} - 4m^2 - 2m - 1}{m^2 + 1} \vee x = \dfrac{\sqrt{- 4m^2 - 12m + 1} + 4m^2 + 2m + 1}{m^2 + 1}

    As stated before the discriminant must be zero:

    -4m^2-12m+1=0~\implies~m=-\dfrac32-\dfrac12 \sqrt{10}~\vee~m=-\dfrac32+\dfrac12 \sqrt{10}

    3. Plug in this value into the equation of the line.

    I've attached a sketch of the circle and the 2 tangents.
    Attached Thumbnails Attached Thumbnails Circle and tangents-zweitang_ankrs.png  
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  3. #3
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    Substitute y+m(x-4) in the equation for the circle. You will get a quadratic equation for x. Write down the condition for this quadratic to have equal roots. That will give you an equation for m.
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  4. #4
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    I have another way if that's OK.

    By completing the square, we find the circle has equation

    (x-1)^{2}+(y-2)^{2}=5

    The tangent line will be a distance from the center of the circle equal to the radius length of \sqrt{5}. The center of the circle is at (1,2)

    Using the distance between a line and a point formula:

    \frac{|-m(1)+1(2)+4m|}{\sqrt{(-m)^{2}+1}}=\sqrt{5}

    Solving for m gives m=\frac{\sqrt{10}-3}{2}, \;\ \frac{-(\sqrt{10}+3)}{2}
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  5. #5
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    Hello, dhiab!

    I think I understand what you're asking . . .


    C is the circle: . x^2 + y^2 - 2x - 4y \:=\:0

    A tangent to C is: . y \:=\:m(x-4)

    \text{Find }m.
    The circle is: . (x-1)^2 + (y-2)^2 \:=\:5

    . . It has center P(1,2) and radius r = \sqrt{5}


    The line . y - 0 \:=\:m(x-4) .has slope m and point Q(0,4)

    Q is one of the y-intercepts of the circle.


    We are asked for the slope of the tangent at Q.


    Code:
                |  /
                | /   * * *
             Q  |/*           *
           (0,4)o               *
               *| *              *
              / |   *
             /* |     * P         *
            / * |       o         *
              * |     (1,2)       *
                |
               *|                *
          - - - * - - - - - - - * - - -
                | *           *
                |     * * *
                |

    The radius PQ has slope: . \frac{4-2}{0-1} \:=\:-2

    The tangent at Q is perpendicular to that radius.

    Therefore: . m \:=\:\frac{1}{2}

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  6. #6
    Super Member dhiab's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, dhiab!

    I think I understand what you're asking . . .

    The circle is: . (x-1)^2 + (y-2)^2 \:=\:5

    . . It has center P(1,2) and radius r = \sqrt{5}


    The line . y - 0 \:=\:m(x-4) .has slope m and point Q(0,4)

    Q is one of the y-intercepts of the circle.


    We are asked for the slope of the tangent at Q.

    Code:
                |  /
                | /   * * *
             Q  |/*           *
           (0,4)o               *
               *| *              *
              / |   *
             /* |     * P         *
            / * |       o         *
              * |     (1,2)       *
                |
               *|                *
          - - - * - - - - - - - * - - -
                | *           *
                |     * * *
                |
    The radius PQ has slope: . \frac{4-2}{0-1} \:=\:-2

    The tangent at Q is perpendicular to that radius.

    Therefore: . m \:=\:\frac{1}{2}
    Hello : THANK YOU
    All lines with the equation pass through the point P(4,0)AND NOT TANGENT IN THE POINT.
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