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    The line with equation mx – y – 2 = 0 touches the circle with equation x2 + 6x +y2 – 8y = 4. Find the two possible values of m, giving your answersin exact form.
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    Firstly, y = mx-2.

    You know they touch e.g. one intersection as the line is tangent to the circle

    You can plug in y=mx-2 and solve for b^2-4ac=0.
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    (Original post by thekidwhogames)
    Firstly, y = mx-2.

    You know they touch e.g. one intersection as the line is tangent to the circle

    You can plug in y=mx-2 and solve for b^2-4ac=0.
    Plug y=mx-2 into the circle equation?
    If yes then how would I solve using b^2-4ac=0??
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    (Original post by Ljm2001)
    Plug y=mx-2 into the circle equation?
    If yes then how would I solve using b^2-4ac=0??
    Plug that in to get an equation in terms of x (this is the equation representing the intersection). As the line is tangent to the circle, b^2-4ac=0. Do this then solve for m.
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    (Original post by Ljm2001)
    The line with equation mx – y – 2 = 0 touches the circle with equation x2 + 6x +y2 – 8y = 4. Find the two possible values of m, giving your answersin exact form.
    to approach this question, you'd need to know one of the basic understandings of quadratic equations, and it's to do with the number of solutions.

    I'd assume you'll know it, but here's a short summation. The number of solutions to a quadratic equation where y = 0 for  ax^2 + bx + c = 0 can be found out using the discriminant. The discriminant carries on from the general equation of a quadratic equation, such that  b^2 - 4ac If u plug i in the values from the quadratic equation to the discriminant, you'll get a number. The number can either be above 0, below 0 or at 0. If it is above 0, then the equation will have 2 real solutions. If it's below 0, it will have no real solution. If it lies on 0, it will have exactly 1 real solution, or known as equal roots.

    Relating this to the original equation, the first step would be to rearrange the equation of the line you've been given in terms of x or y and substitute it into the equation of the circle. This will give u a new quadratic equation which contains a unknown value (other than x or y), and that's generally what you'd have to find.

    HOWEVER, since the circle intersects with the line only once, as it touches the circle, therefore it's a tangent, the number of solutions to the new equation is simply 1. If it had more or less, it would then go through the circle or not touch the circle, respectively.

    Therefore let the discriminant of the new equation equal 0, and from there onwards, find the unknown value.

    Hope this helped.
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    (Original post by Ljm2001)
    Plug y=mx-2 into the circle equation?
    If yes then how would I solve using b^2-4ac=0??
    You'll get a quadratic equation. We then know it is a tangent so the quadratic has equal roots.

    So set the discriminant [b^2 - 4ac] = 0 as this is what we need for equal roots.
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    (Original post by thekidwhogames)
    Plug that in to get an equation in terms of x (this is the equation representing the intersection). As the line is tangent to the circle, b^2-4ac=0. Do this then solve for m.
    Thank you I've worked it out now
 
 
 

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