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Math Help - The Circle?

  1. #1
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    The Circle?

    Hey. Can somebody please help me out with this?

    "A circle passes through the origin (0,0) and the points (4,0) and (3,3).
    Find:
    A. The centre of the circle
    B. The equation to the tangent of the circle to the origin"

    Cheers
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  2. #2
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    Quote Originally Posted by dungo View Post
    "A circle passes through the origin (0,0) and the points (4,0) and (3,3). Find:
    A. The centre of the circle
    B. The equation to the tangent of the circle to the origin"
    Find the perpendicular bisectors of the two line segments with endpoints (0,0)&(4,0) and (0,0)&(3,3).
    Find where those two lines intersect, that is the center of the circle.
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  3. #3
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    Quote Originally Posted by dungo View Post
    Hey. Can somebody please help me out with this?

    "A circle passes through the origin (0,0) and the points (4,0) and (3,3).
    Find:
    A. The centre of the circle
    B. The equation to the tangent of the circle to the origin"

    Cheers
    The general equation of a circle is

    x^2+y^2+2fx+2gy+c=0, where f, g and c are constants.

    Put first point (0,0) in eqn, we got,

    0 + 0 + 0 + 0 + c = 0

    \Rightarrow c = 0

    Now, the eqn becomes,

    x^2+y^2+2fx+2gy=0

    Put second point (4,0) in eqn, we got,

    (4)^2+(0)^2+2f(4)+2g(0)=0

    \Rightarrow f = -2

    Now, the eqn becomes,

    x^2+y^2-4x+2gy=0

    Put third point (3,3) in eqn, we got,

    (3)^2+(3)^2-4(3)+2g(3)=0

    \Rightarrow g = -1

    Now, the eqn becomes,

    x^2+y^2-4x-2y=0

    Complete the square for x and y terms,

    [x^2-2(2)x+2^2]-4+[y^2-2(1)y+1^2]-1=0

    (x-2)^2+(y-1)^2-4-1=0

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

    so, centre = (2, 1) and radius = \sqrt {5}

    B) Now, to find the slope of tangent, we can find the slope of the normal through centre (2, 1) to (0, 0)

    slope of normal = \frac{1-0}{2-0}=\frac{1}{2}

    so, slope of tangent = -2

    now, find the equation of tangent line through (0, 0) having slope -2,

    y = mx + b

    0 = -2(0) + b

    b = 0

    so, eqn of tangent is

    y = -2x

    Please see attached graph
    Attached Thumbnails Attached Thumbnails The Circle?-graph27.jpg  
    Last edited by Shyam; February 21st 2009 at 04:42 PM.
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  4. #4
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    Hello, dungo!

    Did you make a sketch?


    A circle passes through the origin (0,0) and the points A(4,0) and B(3,3).

    Find:
    A. The centre of the circle
    B. The equation to the tangent of the circle to the origin
    Code:
            |       : B
            |       : o (3,3)
            |       *  *
            |     * :   *
            |   *   :    *
            | *     :     *
          - * - - - + - - -o - - -
            |O      :    (4,0)
            |       :      A

    Since O and B are on the x-axis, we already know the perpendicular bisector of OA.
    . . It is the vertical line: x = 2

    Now we seek a point on that vertical line P(2,y)
    . . which is equidistant from O and B

    OP^2 \:=\:2^2 + y^2\:=\:y^2 + 4
    BP^2 \:=\:(2-3)^2 + (y-3)^2 \:=\: y^2 - 6y + 10
    . . We have: . y^2+4 \:=\:y^2 - 6y + 10 \quad\Rightarrow\quad y \:=\:1

    Therefore, the center is (2,1)

    You should be able to complete the problem now . . .

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