Need help on a level maths question I've been stumped on for a while Watch

BlueAgent214
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The question is find the values of m such that y = mx is a tangent to (x-1)^2 + y^2 = 0.8

I tried solving it simultaneously but I only had one answer which isn't right as there are mean to be 2 answers.
I know the centre of the circle is (1,0)
I know the radius is 0.64
But the m is really making life harder.
Can anyone help me?
Pls...
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DennisChin
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(Original post by BlueAgent214)
The question is find the values of m such that y = mx is a tangent to (x-1)^2 + y^2 = 0.8

I tried solving it simultaneously but I only had one answer which isn't right as there are mean to be 2 answers.
I know the centre of the circle is (1,0)
I know the radius is 0.64
But the m is really making life harder.
Can anyone help me?
Pls.
Hi, isn't the radius root 0.8 not 0.8^2?
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DennisChin
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What did you get as your x and y value for solving?
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BlueAgent214
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(Original post by DennisChin)
Hi, isn't the radius root 0.8 not 0.8^2?
Oh yh. Sorry for that mistake
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BlueAgent214
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(Original post by DennisChin)
What did you get as your x and y value for solving?
That's the problem. I don't know how to with the m in the way
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hhhhhhhwusjs
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I'll start you off, draw a radius line (from centre to tangent point).
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simon0
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You have two equations:
1) A curve (in this case a circle):  (x-1)^{2} + y^{2} = 0.8,
2) A straight line:  y = mx.

At the point where the two curves meet, the x and y coordinates must be the same.
One method is to substitute one equation into another.

Spoiler:
Show
Try:
 (x-1)^{2} + \underbrace{(mx)}_{y} \, ^{2} = 0.8.

You should end up with a quadratic you can set to zero.
If you drew the diagram, how many times does the tangent touch the circle?
Then how does this relate to the number of solutions to the quadratic?

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As a separate point, for the radius of the circle, using the equation:

 r^{2} = 0.8.

Then what is the radius?

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Tell us if you need more help.
Last edited by simon0; 3 weeks ago
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BlueAgent214
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Thanks everyone for your help. I got m = 2 and -2.
There are two answers cos there is a power of 2
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DennisChin
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(Original post by BlueAgent214)
The question is find the values of m such that y = mx is a tangent to (x-1)^2 + y^2 = 0.8

I tried solving it simultaneously but I only had one answer which isn't right as there are mean to be 2 answers.
I know the centre of the circle is (1,0)
I know the radius is 0.64
But the m is really making life harder.
Can anyone help me?
Pls...
Wait, isn't there infinite answers to m seeing as a tangent can be at any point on the circle? Unless you're given the values in the equation y=mx
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Pangol
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(Original post by DennisChin)
Wait, isn't there infinite answers to m seeing as a tangent can be at any point on the circle? Unless you're given the values in the equation y=mx
No - the given tangent equation passes through the origin, so a quick sketch shows that there are only two possibilities.
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DennisChin
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(Original post by Pangol)
No - the given tangent equation passes through the origin, so a quick sketch shows that there are only two possibilities.
Ah right yeah that makes sense
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Muttley79
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(Original post by BlueAgent214)
That's the problem. I don't know how to with the m in the way
An easier approach would be to substitute y = mx into the circle - that tells you where the line intersects. Then, because it is a tangent you can look at the discriminant and put that = 0. [two equal roots]
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