Beechey23
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So I got a tricky question that i don't know how to answer can you help me?

The line with equation mx-y-2=0 touches the circle with equation x^2+6x+y^2-8y=4.
Find the two possible values of m
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tnstrk
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Try making y the subject in the equation of the line...
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Beechey23
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(Original post by tnstrk)
Try making y the subject in the equation of the line...
I tried that and I ended up with x^2+6x+m^2x^2-12mx+16=0 and I didn't know where to go after that because there's two unkowns
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mqb2766
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(Original post by Beechey23)
I tried that and I ended up with x^2+6x+m^2x^2-12mx+16=0 and I didn't know where to go after that because there's two unkowns
You want the value(s) of m where there is a single solution to the quadratic equation in x. If there are no solutions, the line does not cut the circle. If there are two solutions, the line cuts the circle. If there is one solution, the line is a tangent to the circle.

So write the quadratic in terms of x, then use the quadratic formula and find the values of m where the discriminant is zero.
Last edited by mqb2766; 1 month ago
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Beechey23
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(Original post by mqb2766)
You want the value(s) of m where there is a single solution to the quadratic equation in x. If there are no solutions, the line does not cut the circle. If there are two solutions, the line cuts the circle. If there is one solution, the line is a tangent to the circle.

So write the quadratic in terms of x, then use the quadratic formula and find the values of m where the discriminant is zero.
Sorry but I don't really understand what you mean by that
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mqb2766
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(Original post by Beechey23)
Sorry but I don't really understand what you mean by that
Just sketch the differerent scenarios. The line could
* miss the circle,
* graze the circumference (tangent)
* cut the circle and intersect the circumference in two places

For the quadratic in x you have, this corresponds to 0, 1 and 2 solutions. A quadratic has 1 solution (repeated) when the discriminant is zero. So write down the discriminant and set it equal to zero to find the values of m which give lines which are tangent to the circle.
Last edited by mqb2766; 1 month ago
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Beechey23
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(Original post by mqb2766)
Just sketch the differerent scenarios. The line could
* miss the circle,
* graze the circumference (tangent)
* cut the circle and intersect the circumference in two places

For the quadratic in x you have, this corresponds to 0, 1 and 2 solutions. A quadratic has 1 solution (repeated) when the discriminant is zero. So write down the discriminant and set it equal to zero to find the values of m which give lines which are tangent to the circle.
It's the part about putting x^2+6x+m^2x^2-12mx+16=0 into a quadratic that's Confusing me because I dont see how you would do it
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tnstrk
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(Original post by Beechey23)
I tried that and I ended up with x^2+6x+m^2x^2-12mx+16=0 and I didn't know where to go after that because there's two unkowns
You may wish to group like terms..
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Beechey23
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(Original post by tnstrk)
You may wish to group like terms..
They are grouped
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mqb2766
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(Original post by Beechey23)
It's the part about putting x^2+6x+m^2x^2-12mx+16=0 into a quadratic that's Confusing me because I dont see how you would do it
(some of the) coefficients depend on m.
x^2*() + x*() + () =
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Beechey23
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(Original post by mqb2766)
(some of the) coefficients depend on m.
x^2*() + x*() + () =
Right okay I understand now that makes alot more sense
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