# Equation of tangent to circle- HELP URGENTLY NEEDED

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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?

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?

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#2

(Original post by

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?

**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?

Could you post your full working I can help you from what you've already done?

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#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

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#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

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

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#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

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#6

**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?

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#7

(Original post by

Where have you got this question from?

**Rolls_Reus_0wner**)Where have you got this question from?

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#8

(Original post by

It is from one of the Edexel new spec maths end of chapter tests. i think only teachers are able to get them.

**number0**)It is from one of the Edexel new spec maths end of chapter tests. i think only teachers are able to get them.

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#9

(Original post by

makes sense cause that came up in my mock. Was proper hard. Need to know where the school gets such hard questions

**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

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#10

(Original post by

Where have you got this question from?

**Rolls_Reus_0wner**)Where have you got this question from?

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#12

(Original post by

Which one?

**Rolls_Reus_0wner**)Which one?

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#15

(Original post by

jheez is that you yeah

**thotproduct**)jheez is that you yeah

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#16

Think there are similar questions in the textbook but this one specifically was in one of the tests I think

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#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

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#18

(Original post by

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

**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

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.

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#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

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

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#20

(Original post by

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

**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

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

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

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