A Level Maths - P3 Coordinate Geometry - Help!

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Emz
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Help! I'm stuck on this question, its from the P3 Edexcel textbook if anyone has it, page 70, Exercise 3A question 11. I'm not sure where to even start this!

The line with equation y=mx is a tangent to the circle with equation
x^2 + y^2 - 6x - 6y + 17 = 0.
Find the possible values of m.

If anyone could point me in the right direction with this it would be great, thanks!

Emma xxx
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JamesF
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Differentiate the equation of the circle, then sub in dy/dx = m
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lgs98jonee
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(Original post by Emz)
Help! I'm stuck on this question, its from the P3 Edexcel textbook if anyone has it, page 70, Exercise 3A question 11. I'm not sure where to even start this!

The line with equation y=mx is a tangent to the circle with equation
x^2 + y^2 - 6x - 6y + 17 = 0.
Find the possible values of m.

If anyone could point me in the right direction with this it would be great, thanks!

Emma xxx
wot so dy/dx=2x+y^2-6-6y=0??
so m=2x+y^2-6-6y??/
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lgs98jonee
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(Original post by JamesF)
Differentiate the equation of the circle, then sub in dy/dx = m
ur a genius? u wanna go to cambridge uni next year?
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elpaw
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(Original post by lgs98jonee)
wot so dy/dx=2x+y^2-6-6y=0??
you have to use implicit differentiation.

dy/dx = -(x-3)/(y-3)
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Emz
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Thank guys! I'll give it a go.

Emma xxx
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JamesF
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(Original post by lgs98jonee)
ur a genius? u wanna go to cambridge uni next year?
Thanks. Yes I would, but its not easy to get in.
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Emz
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Ok, so I differentiated and also got dy/dx = -(x-3) / (y-3)

Now how do I get the values for m from that? (sorry if thats a stupid question and I'm missing something obvious lol)

The answer in the back of the book is (9±√17) / 8 although sometimes they leave answers in weird ways.

Emma xxx
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lgs98jonee
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(Original post by elpaw)
you have to use implicit differentiation.

dy/dx = -(x-3)/(y-3)
oh right...do u learn that in p3 then?
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Emz
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Yeah we do implicit differentiation in P3. Any help with the rest of the question anyone?

Emma xxx
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elpaw
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(Original post by Emz)
Ok, so I differentiated and also got dy/dx = -(x-3) / (y-3)

Now how do I get the values for m from that? (sorry if thats a stupid question and I'm missing something obvious lol)

The answer in the back of the book is (9±√17) / 8 although sometimes they leave answers in weird ways.

Emma xxx
the equation of the tangent is y=mx. you substitute this back into the equation of the circle, i.e. x² + m²x² -6x - 6mx +17 = 0. work from there...
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Emz
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I realise this is asking a lot but is there any chance of you running through that substitution and rearranging for me so I can see how it works?

*Edit - leave that for a minute, I think its just clicked lol
*2nd Edit - no it didnt, help!!!

Emma xxx
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m:)ckel
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x² + m²x² -6x - 6mx +17 = 0

factorise to give x²(1+m²) -6x(1+m) +17 = 0

use quadratic eqn...in the root you get 36(1+m)² - 68(1+m²)

since it is a tangent, only one root, so equate this to 0, and expand:

32m² - 72m + 32 = 0 => 4m² - 9m + 4 = 0

m = [9 ± √(81-64) ] / 8

Therefore m = (9 ± √17) / 8
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elpaw
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maybe a different approach would be helpful.

start with just y = mx. imagine these lines for different m - you will get lines radially from the centre. some of these will cut the circle, some of them will completely miss it, and 2 of them will just touch it a single point. these are the tangents of the circle that go through the centre.

x² + m²x² -6x - 6mx +17 = 0
=> (x²)m² + (-6x)m + (x²-6x+17) = 0

for y=mx to just touch the circle, there has to be a repeated root of that equation, i.e. b²-4ac = 0

36x² - 4x^4 + 24x² - 68 = 0

x^4 - 15x² -17 = 0

x² = (15 +- root(225+68))/2

discard the negative answer (because you are looking for real solutions).

=> x² = 15/2 + rt(293/4)

use this to get a value for y from the circle equation

then use m=dy/dx=-(x-3)/(y-3).
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Emz
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Thanks so much mockel and elpaw, you've both helped loads. I've done the question now, still not totally sure why I had to do the b² - 4ac bit, I'll have another think about it!

Emma xxx
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username9816
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(Original post by Emz)
Thanks so much mockel and elpaw, you've both helped loads. I've done the question now, still not totally sure why I had to do the b² - 4ac bit, I'll have another think about it!

Emma xxx
b^2 - 4ac is the discriminant for a given equation.

The equation has 2 equal roots when b^2 - 4ac = 0

Now apply to this to your given question, and you many understand Elpaw's method.

Sorry I didn't really read the question, as I have only just finished P2 myself.
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lgs98jonee
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(Original post by bono)
b^2 - 4ac is the discriminant for a given equation.

The equation has 2 equal roots when b^2 - 4ac = 0

Now apply to this to your given question, and you many understand Elpaw's method.

Sorry I didn't really read the question, as I have only just finished P2 myself.
only one solution when b^2-4ac=0
but that is wot u say isnt it :-)
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username9816
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(Original post by lgs98jonee)
only one solution when b^2-4ac=0
but that is wot u say isnt it :-)
Hehe, well what I meant was when you find the solution, whether it is + 0 or - 0 will obviously give the same solution.

Sorry, I did really mean 1 solution, but 2 ways of getting that solution when b^2 - 4ac = 0
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