So I managed this question, but I don't the method I used was very effiecent

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smaiu
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Size:  64.3 KBBasically for part C i subed in the y= part, and managed to solve but it took a really long time.
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STEM.lover
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What are your workings?
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DFranklin
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(Original post by smaiu)
Name:  mathsw.JPG
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Size:  64.3 KBBasically for part C i subed in the y= part, and managed to solve but it took a really long time.
You don't really need to solve; use the fact that C is a circle and some basic geometry.
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smaiu
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I literally subbed in y=2x+k into my circle, and formed a quadratic, I then used the discriminant, and made it =0, and found the two values of k. But the mark scheme showed a more efficient way that I don't understand, could you tell me what you'd do to solve this question.
(Original post by STEM.lover)
What are your workings?
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smaiu
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My geometry is not the best, could you tell me where to start
(Original post by DFranklin)
You don't really need to solve; use the fact that C is a circle and some basic geometry.
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(Original post by smaiu)
My geometry is not the best, could you tell me where to start
The other tangent line is going to touch the circle at the point Q that's opposite P

Method 1: Note \vec{PA} = \vec{AQ}.

If you have P and A, that's enough to make it easy to find Q, (I'm not sure what you've worked out already).

Method 2: You should be able to write down a vector in the direction perpendicular to the first tangent (check that it would be going "towards" the circle and not away). If the circle has radius r, the point Q is a distance r from A in the direction of this vector.
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smaiu
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Ok, I understand what you're saying, but can I just throw this in, if the gradient of both lines weren't both 2 would I still be able to do this question through geometry or would I have to sub in, and also is the same gradient the reason why I can assume that PA=AQ
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Anonymouspsych
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Note that because both of the tangent lines have the same gradient, this means that the perpendicular line to both of the tangents must indeed be the same line that crosses through the centre of the circle with gradient -1/2 .

Hence you can work out the coordinates of the point where the 2nd tangent touches the circle using basic vectors.

And finally substitute these coordinate values to find the value of k.
(Original post by smaiu)
My geometry is not the best, could you tell me where to start
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If they were different gradients it would be slightly more complicated but you still wouldn't have to substitute the equation of the tangent line into the equation of the circle. Instead you could find the equation of the line perpendicular tangent and solve the perpendicular and tangent lines simultaneously to work out the coordinates of where the tangent line touches the circle.

Finally you can then substitute the coordinate values in the equation of the tangent to work out the value of k.
(Original post by smaiu)
Ok, I understand what you're saying, but can I just throw this in, if the gradient of both lines weren't both 2 would I still be able to do this question through geometry or would I have to sub in, and also is the same gradient the reason why I can assume that PA=AQ
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