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    How the heck did they get this for (iv) and why did they use a discriminant?
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    And I also need help with the last bit of this question: (iii), how did they find the co-ordinate of the parallel line to the tangent

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    Hi, for Q12 iv you are asked to find the values of k for which the line y=2x+k is a tangent, for the line to be a tangent it must only touch the circle once hence there is only one solution to the equation from part iii. As you know there is only one solution you can make b^2-4ac = 0 as the discriminant will only equal 0 when there is one solution to an equation ( think about the quadratic formula, you will only get one answer from an equation if you are square rooting 0, if you are square rooting any positive number greater than 0 you will get two solutions. ) So then you sub in the values for a,b and c from the equation into b^2-4ac= 0 and solve for k.
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    (Original post by haarithiop)
    Hi, for Q12 iv you are asked to find the values of k for which the line y=2x+k is a tangent, for the line to be a tangent it must only touch the circle once hence there is only one solution to the equation from part iii. As you know there is only one solution you can make b^2-4ac = 0 as the discriminant will only equal 0 when there is one solution to an equation ( think about the quadratic formula, you will only get one answer from an equation if you are square rooting 0, if you are square rooting any positive number greater than 0 you will get two solutions. ) So then you sub in the values for a,b and c from the equation into b^2-4ac= 0 and solve for k.
    Thanks, do you know about how they did part (iii) for question 13"? Like, how they found the co-ordinates of the tangent?
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    (Original post by CorpusLuteum)
    Thanks, do you know about how they did part (iii) for question 13"? Like, how they found the co-ordinates of the tangent?
    parallel tangents on a circle are connected by a diameter. so if you know the coordinates of one of the pair, you just follow the diameter to the other one.
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    (Original post by the bear)
    parallel tangents on a circle are connected by a diameter. so if you know the coordinates of one of the pair, you just follow the diameter to the other one.
    Ooooh, so would you use pythagorean triplet triangles to find the other pair of co-ordinates?
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    (Original post by CorpusLuteum)
    Ooooh, so would you use pythagorean triplet triangles to find the other pair of co-ordinates?
    no it is much easier.

    since the centre is ( 4, 2 ) and one tangent is at ( 4 + √2 , 2 + √2 ) you should be able to see by looking where the other tangent is...
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    (Original post by the bear)
    no it is much easier.

    since the centre is ( 4, 2 ) and one tangent is at ( 4 + √2 , 2 + √2 ) you should be able to see by looking where the other tangent is...
    Oh okay.
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    (Original post by CorpusLuteum)
    Ooooh, so would you use pythagorean triplet triangles to find the other pair of co-ordinates?
    The way I did it was I said that the place where the first tangent meets is 2sqrt(2) up and 2sqrt(2) to the right of the centre of the circle. So therefore the place where the second tangent meets it 2sqrt(2) down and 2sqrt(2) to the left of the centre of the circle so it's (4-2sqrt(2), 2-2sqrt(2))
 
 
 
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