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Math Help - Locus of circumcentre?

  1. #1
    Super Member fardeen_gen's Avatar
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    Locus of circumcentre?

    A triangle has two of its sides along the axes. Its 3rd side touches the circle x^2 + y^2 - 2ax - 2ay + a^2 = 0. Find the equation of locus of the circumcentre of the triangle.
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    A triangle has two of its sides along the axes. Its 3rd side touches the circle x^2 + y^2 - 2ax - 2ay + a^2 = 0. Find the equation of locus of the circumcentre of the triangle.
    I can't provide you with the equation of the locus but I can give you a few hints:

    1.
    x^2 + y^2 - 2ax - 2ay + a^2 = 0~\implies~(x-a)^2+(y-a)^2=a^2
    The circle touches both coordinate axes.

    2. You are dealing with a right triangle where the circumcentre is the midpoint of the hypotenuse.

    3. If the tangentpoint is T(t, p) then the tangent has the equation:

    (x-a)(t-a)+(y-a)(p-a)=a^2

    4. If the tangent is parallel to the coordinate axes then there doesn't exist a circumcentre. Therefore the straight lines x = a and y = a must be asymptotes of the locus.

    5. I've drawn a sketch: Triangles in blue, locus in red, asymptotes in green
    Attached Thumbnails Attached Thumbnails Locus of circumcentre?-ortskrv_umkrsmittelpkte.png  
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  3. #3
    Senior Member pankaj's Avatar
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    Let the circumcentre of the triangle P(x_{1},y_{1}).Therefore the the three vertices of the triangle will be O(0,0),A(2x_{1},0) and B(0,2y_{1}).The Incentre of the triangle will be obviously I(a,a) as can be seen in the diagram supplied by earboth.
    Thus

     <br />
a=\frac{(0)(2\sqrt{x_{1}^2+y_{1}^2})+(0)(2x_{1})+(  2x_{1})(2y_{1})}{2x_{1}+2y_{1}+2\sqrt{x_{1}^2+y_{1  }^2}}<br />


    Recall that the coordinates of the incentre of the triangle having vertices as A(x_{1},y_{1}),B(x_{2},y_{2}) and C(x_{3},y_{3}) is given by
     <br />
(\frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c},\frac{ay_{1}+b  y_{2}+cy_{3}}{a+b+c})<br />

    where a=BC,b=CA,c=AB

    On simplification,we get

     <br />
x_{1}+y_{1}+\sqrt{x_{1}^2+y_{1}^2}=\frac{2x_{1}y_{  1}}{a}<br />

    Therefore the required locus is

     <br />
x+y+\sqrt{x^2+y^2}=\frac{2xy}{a}<br />

    This is the equation of the curve which earboth has drawn in red.
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