# Equations of two tangents to a circle - A Level Maths

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Hi,

Would someone be able to help with the following maths question?

A circle has equation x^2+y^2+4x+12y=-23

The lines l1 and l2 are both tangents to the circle, and they intersect at the point (5,0).

Find the equations of l1 and l2, giving your answers in the form y=mx+c.

Thanks in advance.

Would someone be able to help with the following maths question?

A circle has equation x^2+y^2+4x+12y=-23

The lines l1 and l2 are both tangents to the circle, and they intersect at the point (5,0).

Find the equations of l1 and l2, giving your answers in the form y=mx+c.

Thanks in advance.

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#3

(Original post by

Here is a diagram for the question.

**flappertard**)Here is a diagram for the question.

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(Original post by

What have you worked out so far?

**laurawatt**)What have you worked out so far?

Therefore the centre is (-2,-6) and the radius is √17

Using √(x1-x2)^2+(y1-y2)^2

I can find the length of the line from (5,0) to the centre which is √85

As the tangents meet the circle at 90 degrees you can use Pythagoras to work out the length of the tangent (from (5,0) to where it meets the circle) which is 2√17

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#5

(Original post by

I know that the circle equation in the form (x-a)^2+(y-b)^2=r^2 is (x+2)^2+(y+6)^2=17

Therefore the centre is (-2,-6) and the radius is √17

Using √(x1-x2)^2+(y1-y2)^2

I can find the length of the line from (5,0) to the centre which is √85

As the tangents meet the circle at 90 degrees you can use Pythagoras to work out the length of the tangent (from (5,0) to where it meets the circle) which is 2√17

**flappertard**)I know that the circle equation in the form (x-a)^2+(y-b)^2=r^2 is (x+2)^2+(y+6)^2=17

Therefore the centre is (-2,-6) and the radius is √17

Using √(x1-x2)^2+(y1-y2)^2

I can find the length of the line from (5,0) to the centre which is √85

As the tangents meet the circle at 90 degrees you can use Pythagoras to work out the length of the tangent (from (5,0) to where it meets the circle) which is 2√17

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(Original post by

You've pretty much got most of it. What are you stuck on?

**mqb2766**)You've pretty much got most of it. What are you stuck on?

I don't have the mark scheme but through a graphing tool, I think the points of intersection are (-3,-2) and (1.8,-7.6). However, I need to find this algebraically.

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#7

You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.

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(Original post by

You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.

**mqb2766**)You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.

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#9

(Original post by

Thanks for your help so far. Would you be able to explain this method further? Thank you

**flappertard**)Thanks for your help so far. Would you be able to explain this method further? Thank you

y = m(x-5)

Sub that into the circle to get a quadratic in x. This must have a single solution if the line is a tangent, so solve the discriminant=0 and that should be a quadratic in m, giving the two gradients.

If you wanted to keep going with your approach, you could have got the gradients from the tan(x+y) formula with a few extra lines.

Last edited by mqb2766; 1 month ago

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(Original post by

Your circle equation is a quadratic and your line is

y = m(x-5)

Sub that into the circle to get a quadratic in x. This must have a single solution if the line is a tangent, so solve the discriminant=0 and that should be a quadratic in m, giving the two gradients.

If you wanted to keep going with your approach, you could have got the gradients from the tan(x+y) formula with a few extra lines.

**mqb2766**)Your circle equation is a quadratic and your line is

y = m(x-5)

Sub that into the circle to get a quadratic in x. This must have a single solution if the line is a tangent, so solve the discriminant=0 and that should be a quadratic in m, giving the two gradients.

If you wanted to keep going with your approach, you could have got the gradients from the tan(x+y) formula with a few extra lines.

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#11

(Original post by

Hi, I just used the quadratic discriminant method and I ended up getting a polynomial to the power of 4! Upon solving for m I got 0.22 and -0.58. Any ideas where I am going wrong?

**flappertard**)Hi, I just used the quadratic discriminant method and I ended up getting a polynomial to the power of 4! Upon solving for m I got 0.22 and -0.58. Any ideas where I am going wrong?

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(Original post by

Can you upload your working? You should get a quadratic in x,m

**mqb2766**)Can you upload your working? You should get a quadratic in x,m

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#13

(Original post by

This is my working.

**flappertard**)This is my working.

Can you do it using the tan(x+y) formula? You've got tan(x) and tan(y).

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(Original post by

Will have a look, but it usually falls out a bit easier.

Can you do it using the tan(x+y) formula? You've got tan(x) and tan(y).

**mqb2766**)Will have a look, but it usually falls out a bit easier.

Can you do it using the tan(x+y) formula? You've got tan(x) and tan(y).

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#15

(Original post by

Hi, I've never come across this formula before so not sure what to do. When you say x and y are you referring to the point where the tangent meets the circle? Thanks

**flappertard**)Hi, I've never come across this formula before so not sure what to do. When you say x and y are you referring to the point where the tangent meets the circle? Thanks

Will upload the other one in a few mins.

As you'd worked out these lengths already, it's far simpler to get m this way. The grads are

tan (x+/-y)

Last edited by mqb2766; 1 month ago

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#16

(Original post by

This is my working.

**flappertard**)This is my working.

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