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Math Help - Circles- tangents and equations

  1. #1
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    Circles- tangents and equations

    A circle has the equation x^2 + 8x + y^2 + a = 0. A straight line has the equation y= 3^(1/2)x. Find the value of a for which the line is a tangent to the circle.

    I thought you might have to use the b^2 - 4ac formula because of the tangent to the circle but I just got stuck.
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  2. #2
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    By y= 3^(1/2)x
    I suppose you mean y  = 3 - 1 / 2 x
    cause otherwise it wouldn't be a line equation.

    Well, you just replace the expression for y into the circle equation, which gives

    x^2 + 8x + (3 - 0.5x)^2 + a = 0

    After a little bit of manipulation, you will find a 2nd degree polynomial equation in the form

    A x^2 + B x + C = 0.

    at which point you will - indeed - compute the discriminant D = B^2 - 4 A C, which will be an expression with variable a.
    (e.g. : 14 a - 3.5)

    Discussion:

    • if D > 0 there are 2 real solutions for x - i.e. 2 intersection points between the line and the circle
    • if D<0 there are no real solutions for x - i.e. the line does not 'meet' the circle
    • if D=0 - 1 real solution - the line is tangent to the circle

    So you just solve D=0 to find a.
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  3. #3
    MHF Contributor

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    I would have assumed "y= 3^(1/2)x" mean y= \sqrt{3}x. If that is the case then it is a line through the origin with slope \sqrt{3}. Putting that into the equation of the circle, as Hardwarista suggests, we get x^2+ 8x+ 3x^2+ a= 0 or 4x^2+ 8x+ a= 0 which we can also write as x^2+ 2x+ a/4= 0. If the line is tangent to the circle, this equation must have a single solution so the left side must be a perfect square. Since (x+1)^2= x^2+ 2x+ 1, it's pretty easy to solve for a.
    Last edited by Moo; March 7th 2009 at 07:59 AM. Reason: math tag mistake ;)
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