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Posted by on Wednesday, December 28, 2011 at 12:33am.

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).

  • algebra - , Wednesday, December 28, 2011 at 4:32am

    The center of the circle also lies on the perpendicular bisector of any chord. We have the chord between the two given points.
    The chord has slope (0-4)/(5-1) = -1.
    The midpoint of the chord is (3,2).
    The equation of the line containing (3,2) and perpendicular to the chord is

    (y-2)/(x-3) = 1
    or
    -x + y = -1
    This intersects the line
    x + y = 3 at
    (2,1)

    So, our circle has equation

    (x-2)^2 + (y-1)^2 = r^2
    plug in either of the given points and we find r^2 = 10. so,

    (x-2)^2 + (y-1)^2 = 10
    x = 2 + √10 cos t
    y = 1 + √10 sin t

    the rest should be easy to figure out.

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