# A Level Maths - P3 Coordinate Geometry - Help!

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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

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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#3

(Original post by

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

**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

so m=2x+y^2-6-6y??/

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#4

(Original post by

Differentiate the equation of the circle, then sub in dy/dx = m

**JamesF**)Differentiate the equation of the circle, then sub in dy/dx = m

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#5

(Original post by

wot so dy/dx=2x+y^2-6-6y=0??

**lgs98jonee**)wot so dy/dx=2x+y^2-6-6y=0??

dy/dx = -(x-3)/(y-3)

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#7

(Original post by

ur a genius? u wanna go to cambridge uni next year?

**lgs98jonee**)ur a genius? u wanna go to cambridge uni next year?

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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

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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#9

(Original post by

you have to use implicit differentiation.

dy/dx = -(x-3)/(y-3)

**elpaw**)you have to use implicit differentiation.

dy/dx = -(x-3)/(y-3)

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Yeah we do implicit differentiation in P3. Any help with the rest of the question anyone?

Emma xxx

Emma xxx

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#11

(Original post by

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

**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

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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

*Edit - leave that for a minute, I think its just clicked lol

*2nd Edit - no it didnt, help!!!

Emma xxx

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#13

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

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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#14

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).

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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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

Emma xxx

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#16

(Original post by

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

**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

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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#17

(Original post by

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.

**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.

but that is wot u say isnt it :-)

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#18

(Original post by

only one solution when b^2-4ac=0

but that is wot u say isnt it :-)

**lgs98jonee**)only one solution when b^2-4ac=0

but that is wot u say isnt it :-)

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

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