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

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#1
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.

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#2
Here is a diagram for the question.
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1 month ago
#3
(Original post by flappertard)
Here is a diagram for the question.
What have you worked out so far?
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#4
(Original post by laurawatt)
What have you worked out so far?
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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1 month ago
#5
(Original post by 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
You've pretty much got most of it. What are you stuck on?
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#6
(Original post by mqb2766)
You've pretty much got most of it. What are you stuck on?
The question is asking for the equations of the tangents. Not quite sure how to get there.

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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1 month ago
#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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#8
(Original post by mqb2766)
You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.
Thanks for your help so far. Would you be able to explain this method further? Thank you
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1 month ago
#9
(Original post by 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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#10
(Original post by mqb2766)
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.
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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1 month ago
#11
(Original post by 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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#12
(Original post by mqb2766)
This is my working.
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1 month ago
#13
(Original post by flappertard)
This is my working.
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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#14
(Original post by 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).
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
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1 month ago
#15
(Original post by 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
For the circle - line - tangent one, I'd transform the origin to the point (5,0). That simplified the calculation a lot as the line is y=mx. The gradient is unchanged.

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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1 month ago
#16
(Original post by flappertard)
This is my working.
The coeff. of is incorrect in your quadratic - should be 0
X

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