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March 27, 2017

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Given that the equation of the circle is x^2+y^2+2x-8y+5=0
(i)find the centre and radius of the circle
(ii)Determine whether the line x-3y=8 is tangent to the given circle

  • maths - ,

    x^2+2x+1 + y^2-8y+16=1+16-5
    (x+1)^1 + (y-4)^2=12

    center, -1,4
    radius sqrt12

  • maths - ,

    complete the square ....

    x^2 + 2x + .... + y^2 - 8y + ... = -5
    x^2 + 2x + 1 + y^2 - 8y + 16 = -5+1+16
    (x+1)^2 + (y-4)^2 = 12

    centre is (-1,4) , radius is √12 or 2√3

    if x-3y=8 is a tangent, then the distance from (-1,4) to the line should be √12

    distance from (-1,4) to x-3y-8=0 is
    |-1 - 3(4) - 8|/√(1+9)
    = 21/√10 ≠ √12
    so, no, it is not a tangent.

    other way, find intersection....
    x = 3y+8 , sub into circle

    (3y+8)^2 + y^2 + 2(3y+8) -8y+5=0
    9y^2 + 48y + 64 + y^2 + 6y + 24 - 8y + 5 = 0
    10y^2 +46y +93=0

    to be tangent, this quadratic should have one solution, thus it should be a perfect square.
    But it is not, so .... no tangent

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