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posted by on .

can u help
this the qustion is :

A circle passes through the points (1,4)&(5,0) and its centre lies on the line x+y-3=0.Find
(i) the equation of the circle and it parametric equations .
(ii) the centre,diameter and area of the circle .
(iii) the equation of the tangent at (1,4) .
(iv) the equation of the circle concentric with the above circle and passing through the point (7,8).

  • math - ,

    let A be (1,4) and B be (5,0)
    So AB must be a chord of the circle and the centre must be the intersection of the right-bisector of AB and the line x+y - 3 = 0
    slope of AB = -4/4 = -1
    slope of right-bisector = 1
    midpoint of AB = (3,2)
    equation of right-bisector: y = x + b
    with (3,2) on it
    2 = 3 + b ---> b = -1
    right-bisector: y = x-1

    solve with the second equation:
    x+y - 3 = 0
    x + (x-1) = 3
    2x = 4
    x = 2, then y = 3
    centre is (2,3)
    radius is distance from centre to A
    = √((2-1)^2 + (3-4)^2) = √2

    i) equation of circle: (x-2)^2 + (y-3)^2 = 2
    ii) you do it
    iii) slope of radius from A to centre = ...
    slope of tangent at A is ..... (the negative reciprocal of above
    now you have a point and slope of the tangent....
    iv) equation will be
    (x-2)^2 + (y-3)^ = r^2
    plug in (7,8) to get r^2

  • ARghh ---math - ,

    Just noticed that you posted the same question twice, and that both Steve and I answered it.
    BUT, I made a stupid error in my centre calculation, (that's what you get when you try to do things in your head without writing them down)

    I had the centre as (2,3), should have been (2,1) like Steve had
    so go with his solution.

  • math - to Reiny - ,

    Heh heh -- We all do these problems just for fun anyway. Any help to the students is just a welcome side effect. And, these little boo-boos help keep us humble...

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