A Level Maths - P3 Coordinate Geometry - Help!

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

Reply 2

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??/

Reply 3

JamesF

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

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

Reply 4

elpaw

15

lgs98jonee

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

you have to use implicit differentiation.

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

Reply 5

OP

Thank guys! I'll give it a go.

Emma xxx

Reply 6

JamesF

lgs98jonee

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

Thanks. Yes I would, but its not easy to get in.

Reply 7

OP

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

Reply 8

elpaw

you have to use implicit differentiation.

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

oh right...do u learn that in p3 then?

Reply 9

OP

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

Emma xxx

Reply 10

elpaw

15

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...

Reply 11

OP

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

Reply 12

m:)ckel

14

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

Reply 13

elpaw

15

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).

Reply 14

OP

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

Reply 15

username9816

14

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. :smile:

Sorry I didn't really read the question, as I have only just finished P2 myself.

Reply 16