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Math Help - locus of mid point of line inersecting circle

  1. #1
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    locus of mid point of line inersecting circle

    A circle passes through the points A, B, and C which have coordinates (0,3), (\sqrt{3},0), (-\sqrt{3},0) respectively.
    Equation of circle x^2+y^2-2y=3
    A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM.
    I found the values of x and y in terms of m.
    (\frac{4m+2\sqrt{m^2-3}}{m^2+1}, \frac{m^2+2m\sqrt{m^2-3}-3}{m^2+1})  (\frac{4m-2\sqrt{m^2-3}}{m^2+1}, \frac{m^2-2m\sqrt{m^2-3}-3}{m^2+1})
    Now what's next? I don't know hot to proceed from here. Any help is greatly appreciated! Thanks!
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  2. #2
    Senior Member pacman's Avatar
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    "A circle passes through the points A, B, and C which have coordinates respectively. Equation of circle
    A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM."

    How i do it . . . .

    x^2 + y^2 - 2y = 3 - - - - (1)

    and y = mx + 3 - - - - (2) , substituting
    _________________________________________

    x^2 + (mx + 3)^2 - 2(mx + 3) = 3,

    x^2 + m^2x^2 + 6mx + 9 - 2mx - 6 = 3,

    (1 + m^2)x^2 + 4mx = 3 - 9 + 6 = 0,

    (1 + m^2)x^2 + 4mx = 0

    x[(1 + m^2)x + 4m] = 0,

    x = 0, then y = 3

    x = -4m/(1 + m^2), then y = (3 - m^2)/(1 + m^2)

    i don't know what to continue either . . ..
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  3. #3
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    Hello arze
    Quote Originally Posted by arze View Post
    A circle passes through the points A, B, and C which have coordinates (0,3), (\sqrt{3},0), (-\sqrt{3},0) respectively.
    Equation of circle x^2+y^2-2y=3
    A line y=mx-3of variable gradient m, cuts the circle at L and M. Find the Cartesian equation of the locus of the midpoint of the line LM.
    I found the values of x and y in terms of m.
    (\frac{4m+2\sqrt{m^2-3}}{m^2+1}, \frac{m^2+2m\sqrt{m^2-3}-3}{m^2+1})  (\frac{4m-2\sqrt{m^2-3}}{m^2+1}, \frac{m^2-2m\sqrt{m^2-3}-3}{m^2+1})
    Now what's next? I don't know hot to proceed from here. Any help is greatly appreciated! Thanks!
    The straight line y=mx-3 meets x^2+y^2-2y=3 where

    x^2 +(mx-3)^2 -2(mx-3)=3

    \Rightarrow x^2(1+m^2)-8mx+12=0

    If the roots of this equation are x_1,x_2, then x_1+x_2 = \frac{8m}{1+m^2}

    and if the line meets the circle where y = y_1,y_2, then y_1+y_2 = mx_1-3+mx_2 - 3 = m(x_1+x_2)-6=\frac{8m^2}{1+m^2}-6=\frac{2m^2-6}{1+m^2}

    Now the mid-point of the line joining (x_1,y_1), (x_2,y_2) is \Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big). So its locus is given by:

    x=\frac{4m}{1+m^2} (1)

    y = \frac{m^2-3}{1+m^2} (2)

    From (2): m^2 = \frac{3+y}{1-y} (3)

    and from (1) and (2): 1+m^2 = \frac{4m}{x}=\frac{m^2-3}{y}

    Subst from (3):

    \frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{1-y}

    \Rightarrow \frac{3+y}{x^2(1-y)}=\frac{y^2}{(1-y)^2}

    \Rightarrow x^2y^2 = (3+y)(1-y)

    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello arzeThe straight line y=mx-3 meets x^2+y^2-2y=3 where

    x^2 +(mx-3)^2 -2(mx-3)=3

    \Rightarrow x^2(1+m^2)-8mx+12=0

    If the roots of this equation are x_1,x_2, then x_1+x_2 = \frac{8m}{1+m^2}

    and if the line meets the circle where y = y_1,y_2, then y_1+y_2 = mx_1-3+mx_2 - 3 = m(x_1+x_2)-6=\frac{8m^2}{1+m^2}-6=\frac{2m^2-6}{1+m^2}

    Now the mid-point of the line joining (x_1,y_1), (x_2,y_2) is \Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big). So its locus is given by:

    x=\frac{4m}{1+m^2} (1)

    y = \frac{m^2-3}{1+m^2} (2)

    From (2): m^2 = \frac{3+y}{1-y} (3)

    and from (1) and (2): 1+m^2 = \frac{4m}{x}=\frac{m^2-3}{y}

    Subst from (3):

    \frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{1-y}

    \Rightarrow \frac{3+y}{x^2(1-y)}=\frac{y^2}{(1-y)^2}

    \Rightarrow x^2y^2 = (3+y)(1-y)

    Grandad
    Thanks!
    I can't get the answer though, its supposed to be (x^2+y^2+2y-3)(y+3)=0
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  5. #5
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    Hello arze
    Quote Originally Posted by arze View Post
    Thanks!
    I can't get the answer though, its supposed to be (x^2+y^2+2y-3)(y+3)=0
    Sorry, I lost a term in y just here:

    \frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{1-y}
    It should be:

    \frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{\color{red}y\color{black}(1-y)}=\frac{4}{1-y}

    \Rightarrow \frac{3+y}{x^2(1-y)}=\frac{1}{(1-y)^2}

     \Rightarrow x^2=(3+y)(1-y)

    which is, I think, the correct answer.

    I cannot understand why they have written the answer in the form that you quoted:

    (x^2+y^2+2y-3)(y+3)=0

    This is the product of two factors which equals zero; i.e.

    x^2+y^2+2y-3=0

    or

    y+3=0

    The first of these, when re-arranged, is the same as my revised answer. The second is a horizontal line, which, as far as I can see, does not form part of the locus of the mid-point of the chord.

    Grandad
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  6. #6
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    Quote Originally Posted by Grandad View Post
    Hello arzeSorry, I lost a term in y just here:

    It should be:

    \frac{4}{x}\sqrt{\frac{3+y}{1-y}}= \frac{\dfrac{3+y}{1-y}-3}{y}=\frac{4y}{\color{red}y\color{black}(1-y)}=\frac{4}{1-y}

    \Rightarrow \frac{3+y}{x^2(1-y)}=\frac{1}{(1-y)^2}

     \Rightarrow x^2=(3+y)(1-y)

    which is, I think, the correct answer.

    I cannot understand why they have written the answer in the form that you quoted:

    (x^2+y^2+2y-3)(y+3)=0

    This is the product of two factors which equals zero; i.e.

    x^2+y^2+2y-3=0

    or

    y+3=0

    The first of these, when re-arranged, is the same as my revised answer. The second is a horizontal line, which, as far as I can see, does not form part of the locus of the mid-point of the chord.

    Grandad
    I was quite confused with that also. but it was worth asking. Thanks!
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