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Geometry HELP HELP

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Find the equation of a circle circumscribes a triangle determined by the line y= 0 , y= x and 2x+3y= 10
PLEASE HELP ME BELLS

  • Geometry HELP HELP -

    Make a sketch, and by some simple preliminary algebra you can see that the triangle is
    A(0,0) , B(2,2) and C(0, 10/3)

    the centre of the circle lies on the right bisectors of any of these sides
    midpoint of AC = (0, 5/3) so right-bisector equation is
    y = 5/3
    midpoint of B(2,2) is (1,1)
    slope of AB = 1
    so slope of bisector at (1,1) is =1
    y-1 = -1(x-1)
    x + y = 2
    sub in y = 5/3
    x = 2-5/3 = 1/3

    centre is (1/3 , 5/3)
    equation:
    (x - 1/3)^2 + (y-5/3)^2 = r^2
    plug in (0,0) which lies on it
    1/9 + 25/9 = r^2 = 26/9

    (x-1/3)^2 + (y-5/3)^2 = 26/9

  • back up - Geometry HELP HELP -

    oops, I misread the question, and got the wrong triangle
    (I used the y-axis, instead of the x-axis)

    so thepoints are (0,0), (2,2) and (5,0)

    same equation for the right-bisector of AB which is
    x + y = 2

    the right bisector of line from A to (5,0) is
    x = 5/2

    for centre:
    5/2 + y = 2
    y = -1/2
    centre is (5/2, -1/2)

    (x-5/2)^2 + (y+1/2)^2 = r^2
    for (0,0) , which lies on the circle
    25/4 + 1/4 = r^2 = 26/4

    (x - 5/2)^2 + (y+ 1/2)^2 = 26/4

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