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Posted by on Friday, March 23, 2012 at 5:02am.

the circle is given by the equation x^2+y^2+8x-7/2y=0
find
(i)coordinates of the centre and the radius
(ii)coordinates of the point where the circle crossesthe x-axis

  • maths - , Friday, March 23, 2012 at 5:11am

    oh well

    x^2 + 8 x = -y^2 + 3.5 y

    x^2 + 8 x + 16 = -y^2 + 3.5 y + 16

    (x+4)^2 = -y^2 + 3.5 y + 16

    y^2 - 3.5 y = -(x+4)^2 + 16

    y^2 - 3.5 y + 3.0625 + (x+4)^2 = 19.0625

    (x+4)^2 + (y-1.75)^2 = 4.366^2

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