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Math Help - Coordinate Geometry - equation of the circle

  1. #1
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    Coordinate Geometry - equation of the circle

    I don't know how to start solving this problem.
    I have drawn the graph but have no idea to start.

    A circle C cuts the x-axis at A and B, and the y-axis at P and Q. It is given that AB = 2 and major arc PQ : minor arc PQ = 3:1. The distance from the centre G of the circle C to the line x - 2y = 0 is d = 1 / (5^(1/2)).
    Find the equation of the circle C.

    Can anyone give me some hints?
    Thank in advance.
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  2. #2
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    Hello, ling_c_0202!

    I have an approach, but it's very messy.
    . . I'll leave the algebra up to you . . .


    A circle C cuts the x-axis at A and B, and the y-axis at P and Q.
    It is given that AB = 2 and (major arc PQ):(minor arc PQ = 3:1). . . . . . . . . ._
    The distance from the centre G of the circle C to the line x - 2y = 0 is d = 1/√5
    Find the equation of the circle C.
    Code:
              |     * * *
              | *           *
             Q*               *
             *| *              *
              |   *
            * |     * G         *
            * |       *         *
            * |     ** *        *
              |   * *   *
            P*| *  *     *     *
              *   *       *   *
          ----+--*---------*------
              | A   * * *   B

    The center of the circle is: G(h,k); its radius is r.

    Let the four intercepts be: .A(a,0), B(b,0), P(0,p), Q(0.q)

    We have seven unknowns to solve for
    . . but we have enough equations.


    GA = r: . (h - a) + k .= .r

    BG = r: . (h - b) + k .= .r

    GP = 4: . h + (k - p) .= .r

    GQ = r: . h + (k - q) .= .r


    We know that: .b - a .= .2

    From the ratio of the arcs, we know that: angle PGQ = 90.
    Then: .PQ] .= .r√2 . . . Hence: .q .= .p + r√2

    Since GP is perpendicular to GQ, their slopes are negative reciprocals.
    So that: .m(GP) = (k - p)/h is the negative reciprocal of m(GQ) = (k - q)/h


    I told you it was messy . . . Good luck!

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  3. #3
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    I actually had the same approach as Soroban.
    ---
    Let me add something.
    It seems to me there are 2 possible circles here.
    We are told that the center of the circle is located sqrt(1/5)
    from the line x-2y=0.
    That means the locus of point satisfing the condition that it needs to be located sqrt(1/5) is a pair of parallel line whose distance is sqrt(1/5) from x-2y=0.
    Since the line is parallel its equation is,
    y=(1/2)x+b
    One point on the line is: (0,b)
    The distance from x-2y=0 is sqrt(1/5)
    Thus, by distance formula for a line,
    |-2b+0|/sqrt(1+2^2)=sqrt(1/5)
    Thus,
    |-2b|=1
    Thus,
    b=+/- 1/2
    Thus, there are two possible lines on which the center of circle lies upon.
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