# Equation of tangent to circle- HELP URGENTLY NEEDED

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#1
The line with equation mx-y-2=0 touches the circle with equation x^2+6x+y^2-8y=4. Find 2 possible values of m, giving your answers in exact form.

I've attempted everything, but I can't seem to do it.

I found the centre of the circle and found the gradient of the circle to be -2, making the gradient of the tangent 0.5.

I'm so confused, can someone please explain how to get the answers?
0
4 years ago
#2
(Original post by bananabrainedbee)
The line with equation mx-y-2=0 touches the circle with equation x^2+6x+y^2-8y=4. Find 2 possible values of m, giving your answers in exact form.

I've attempted everything, but I can't seem to do it.

I found the centre of the circle and found the gradient of the circle to be -2, making the gradient of the tangent 0.5.

I'm so confused, can someone please explain how to get the answers?
Hey! Sorry you haven't had a response yet.

0
4 years ago
#3
Find a coordinate at the radius of the circle then use the equation y-y1 = 0.5(x-x1) to get the equation of the line
0
4 years ago
#4
1) Write mx-y-2=0 in the form y=mx+c
2) work out equation of circle
3) Subsitute y=mx-2 into the y in the equation of the circle
4) Expand and simplify
5) factor out x so in brackets you get m and numbers only
6) work out discriminant (b^2-4ac)
7) sub into m
5
4 years ago
#5
I remember doing this question in one of those end of chapter tests, you need to remember how the discriminant is equal to zero when they only touch
0
3 years ago
#6
(Original post by bananabrainedbee)
The line with equation mx-y-2=0 touches the circle with equation x^2+6x+y^2-8y=4. Find 2 possible values of m, giving your answers in exact form.

I've attempted everything, but I can't seem to do it.

I found the centre of the circle and found the gradient of the circle to be -2, making the gradient of the tangent 0.5.

I'm so confused, can someone please explain how to get the answers?
Where have you got this question from?
0
3 years ago
#7
(Original post by Rolls_Reus_0wner)
Where have you got this question from?
It is from one of the Edexel new spec maths end of chapter tests. i think only teachers are able to get them.
0
3 years ago
#8
(Original post by number0)
It is from one of the Edexel new spec maths end of chapter tests. i think only teachers are able to get them.
makes sense cause that came up in my mock. Was proper hard. Need to know where the school gets such hard questions
0
3 years ago
#9
(Original post by Rolls_Reus_0wner)
makes sense cause that came up in my mock. Was proper hard. Need to know where the school gets such hard questions
Yeh, defo set by the exam board. To be fair some textbook questions can be quite difficult
0
3 years ago
#10
(Original post by Rolls_Reus_0wner)
Where have you got this question from?
theyre in the as maths textbook
0
3 years ago
#11
Which one?
0
3 years ago
#12
(Original post by Rolls_Reus_0wner)
Which one?
https://www.amazon.co.uk/Edexcel-lev.../dp/129218339X
0
3 years ago
#13
What page
0
3 years ago
#14
jheez is that you yeah
0
3 years ago
#15
(Original post by thotproduct)
jheez is that you yeah
im out here on that as maths tingg
0
3 years ago
#16
Think there are similar questions in the textbook but this one specifically was in one of the tests I think
0
3 years ago
#17
Would you be able to tell me what you mean by work out the equation of the circle?? Im not sure what to do with it thanks
0
3 years ago
#18
(Original post by kateroseco)
Would you be able to tell me what you mean by work out the equation of the circle?? Im not sure what to do with it thanks
you need to complete the square of the given equation of the square for x and y respectively so the equation of the circle is in the form (x-a)^2 + (y-b)^2 = r^2.

Actually, you don't have to do that. You can simply sub y=mx-2 into the equation of the given circle if you want without putting it in correct circle form . step 1 isn't needed for this question.
1
3 years ago
#19
okay so i got:

x^2(1+m^2) + x(6-12m) + 16 = 0

and im not sure which bits to use for the b^2-4ac.... thank you for your help
0
3 years ago
#20
(Original post by kateroseco)
okay so i got:

x^2(1+m^2) + x(6-12m) + 16 = 0

and im not sure which bits to use for the b^2-4ac.... thank you for your help
Then parts inside the bracket, as you are solving to find m, the unknown. Thats why you took out x, to be able to solve for m.

So you know the discriminant is 0 as the lines touch the circle

(6-12m)^2 - (4)(1+m^2)(16)=0
2
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