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# Have I invented a circle theorem? watch

1. Problem solved I haven't invented a circle theorem. I've used another geometric theorem which I had not really/sort of heard of before, but never really with a circle.
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2. No you haven't.
3. (Original post by RattyVlogger)
No you haven't.
What is it then?
4. no
5. (Original post by AverageN3RD)
no
Then what have I done?
6. lovin that picture on the left XD but yeh what you've done isn't circle theorems, just a variety of algebraic methods like pythagoras' theorem, expansion/factorising quadratics etc. hard to explain as i don't know many smart maths terms to describe but yeh, just a slight twist on the cartesian equation of a circle. relatively surprised the mark scheme didnt notice a method like that as that's not TOO unpredictable. I myself would do something a little similar
oh yeh, also i don't believe you can remove attachments after you add them...
7. You have done the question.

Congratulations! Here, have 4 marks.
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8. (Original post by Toasticide)
lovin that picture on the left XD but yeh what you've done isn't circle theorems, just a variety of algebraic methods like pythagoras' theorem, expansion/factorising quadratics etc. hard to explain as i don't know many smart maths terms to describe but yeh, just a slight twist on the cartesian equation of a circle. relatively surprised the mark scheme didnt notice a method like that as that's not TOO unpredictable. I myself would do something a little similar
oh yeh, also i don't believe you can remove attachments after you add them...
WHAT!!! I don't even know why I have that picture! It's not even me!!! Would you not find the gradient of AB and (-1/m) like a sensible person? Also, can you explain how what I've done can equal the equation of l. What I've done makes sense, but not for that purpose if you see what I mean...
9. (Original post by oniisanitstoobig)
You have done the question.
Congratulations! Here, have 4 marks.
I have, but it is genuinely worth max. 1 mark.
10. (Original post by 04MR17)
I have, but it is genuinely worth max. 1 mark.
I don't know, I personally wouldn't do all that work for just 1 mark.
11. You're going to have to be a bit more descriptive. Where do you think you've used the new 'theorem'?

Your working is incorrect, but the error cancels itself out. You cannot equate the two distances to , but can equate them to each other, since PM bisects AB.
12. (Original post by oniisanitstoobig)
I don't know, I personally wouldn't do all that work for just 1 mark.
No, the question is worth 4 marks, my answer is worth one mark, based on the mark scheme.
13. (Original post by 04MR17)
OK, I was doing a question, and though I don't know how, ended up getting the correct answer, using a completely unconnected method which I don't even know is a method. I was doing Edexcel C2 (exam tomorrow) January 2007 paper (Question 10

I know the answer. And I know the sensible way of working it out. My question is this: is this a thing?
If so: what is it?

EDIT: Please ignore the random picture, I don't know why it is there. Anyone know how to get rid of it?
Why did you work out the coordinates of B? Surely it would have been easier to work out the gradient of AM and then line l is just perpendicular. I can't really understand what you've done, can you explain it?
14. (Original post by 04MR17)
Then what have I done?
You've accidentally come to the correct conclusion despite making an incorrect assumption on the way. You've taken a generic point on the line ( say) with coordinates and then noted that, since lies on the perpendicular bisector of , it must be equidistant from and for every choice of points for

The incorrect bit is that you claim that the distance to this point from and is always 13. Which is clearly nonsense. However, since you equated them both to 13, it turned out OK.

EDIT: As Morgan8002 points out above.
15. (Original post by morgan8002)
You're going to have to be a bit more descriptive. Where do you think you've used the new 'theorem'?

Your working is incorrect, but the error cancels itself out. You cannot equate the two distances to , but can equate them to eat other, since PM bisects AB.
I just titled the thread to get some attention. "What have I done?" was always going to be a little bit vague.
Can you find the equation of a perpendicular bisector of the chord between two points by putting the coordinates of the points into the equation of a circle (without knowing the centre) and equating them?
16. (Original post by 04MR17)
OK, I was doing a question, and though I don't know how, ended up getting the correct answer, using a completely unconnected method which I don't even know is a method. I was doing Edexcel C2 (exam tomorrow) January 2007 paper (Question 10

I know the answer. And I know the sensible way of working it out. My question is this: is this a thing?
If so: what is it?

EDIT: Please ignore the random picture, I don't know why it is there. Anyone know how to get rid of it?
17. (Original post by NickLCFC)
Why did you work out the coordinates of B? Surely it would have been easier to work out the gradient of AM and then line l is just perpendicular. I can't really understand what you've done, can you explain it?
Explained in previous reply. I don't know why I didn't go for the obvious, but let's not dwell on things that didn't happen.
(Original post by Farhan.Hanif93)
You've accidentally come to the correct conclusion despite making an incorrect assumption on the way. You've taken a generic point on the line ( say) with coordinates and then noted that, since lies on the perpendicular bisector of , it must be equidistant from and for every choice of points for
The incorrect bit is that you claim that the distance to this point from and is always 13. Which is clearly nonsense. However, since you equated them both to 13, it turned out OK.
EDIT: As Morgan8002 points out above.
So is it a fluke?
18. (Original post by M14B)
Thank you that was very helpful.
19. (Original post by 04MR17)
I just title the thread to get some attention. "What have I done?" was always going to be a little bit vague.
Can you find the equation of a perpendicular bisector of the chord between two points by putting the coordinates of the points into the equation of a circle (without knowing the centre) and equating them?
Given any point on the bisector, the distances between that point and each of A and B are the same, which is equivalent to A and B lying on a circle centred on that point. You know the point, but it's arbitrary. So yeah. You just don't know the radius, but you know the radius is the same for both points.
20. (Original post by morgan8002)
Given any point on the bisector, the distances between that point and each of A and B are the same, which is equivalent to A and B lying on a circle centred on that point. You know the point, but it's arbitrary. So yeah. You just don't know the radius, but you know the radius is the same for both points.
Just to confirm, are you telling me that this is an actual method?

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