# Have I invented a circle theorem?Watch

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3 years ago
#21
(Original post by 04MR17)
Just to confirm, are you telling me that this is an actual method?
Yes. The only problem with your working was assuming that the distances were .
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#22
(Original post by morgan8002)
Yes. The only problem with your working was assuming that the distances were .
Is this a (well) known method?
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3 years ago
#23
(Original post by 04MR17)
Is this a (well) known method?
Fairly.
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#24
(Original post by morgan8002)
Fairly.
Oh. Well our maths teachers are crap then, they didn't have a clue how I could get the answer from two points not on the line.
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3 years ago
#25
(Original post by 04MR17)
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.So is it a fluke?
Yes and no; it is more of an error that cancels itself out - the flukey bit was that it did.

In case it isn't clear, your answer in fact had nothing to do with circles at all. Instead, merely that the distances of the endpoints of a bisected line are equidistant from each point on the perpendicular bisector. Which isn't news or anything like that, and isn't necessarily an unusual way to find the equation of a perpendicular bisector.
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3 years ago
#26
(Original post by 04MR17)
.
I've removed the random photo for you.
1
3 years ago
#27
(Original post by 04MR17)
Just to confirm, are you telling me that this is an actual method?
I wouldn't call it a method (certainly not a circle theorem), but rather you've just used a well-known fact which is a recurrent theme in questions like this.
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#28
(Original post by oniisanitstoobig)
I wouldn't call it a method (certainly not a circle theorem), but rather you've just used a well-known fact which is a recurrent theme in questions like this.
Could you please state this fact?

(Original post by Farhan.Hanif93)
Yes and no; it is more of an error that cancels itself out - the flukey bit was that it did.
But, ignoring the error. Are you saying that the answer was a fluke?

(Original post by oniisanitstoobig)
In case it isn't clear, your answer in fact had nothing to do with circles at all. Instead, merely that the distances of the endpoints of a bisected line are equidistant from each point on the perpendicular bisector. Which isn't news or anything like that, and isn't necessarily an unusual way to find the equation of a perpendicular bisector.
So did I have to use the equation of a circle? What would you use instead?
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#29
Just found this online:
Perpendicular Bisector TheoremLet's talk perpendicular bisectors. These lines are immensely useful. Let's say you're an architect. I wanted to be an architect once. Then I realized it was less building cool models and more building regulations and codes. Anyway, let's completely ignore building codes here.Here's some land. You're designing a skyscraper. The folks who hired you made two demands: it needs to go straight up and it needs to be in the middle of the land. Seems simple enough, right?Say, do you know what you did? You made a perpendicular bisector. And it's not as simple as it seems. Consider the Leaning Tower of Pisa. Maybe it bisects the plots of land. But it definitely isn't perpendicular. Granted, who would visit the perpendicular tower of Pisa? These folks definitely ignored some building codes.Anyway, perpendicular bisectors come with their very own theorem. The perpendicular bisector theorem states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the segment's endpoints. In other words, if we hung laundry lines from any floor of our tower, each floor would use the same length of laundry line to reach the ground. Okay, but I'm guessing the neighbors might complain about all the underwear hanging outside our tower
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#30
(Original post by oniisanitstoobig)
I wouldn't call it a method (certainly not a circle theorem), but rather you've just used a well-known fact which is a recurrent theme in questions like this.
That fact, is the perpendicular bisector theorem. Quod Erat Demonstrandum.
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3 years ago
#31
(Original post by 04MR17)
That fact, is the perpendicular bisector theorem. Quod Erat Demonstrandum.
You didn't use that fact. You took a different approach.

The fact you used is that the distance from the centre to the circumference is the radius of the circle, for all points.
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#32
(Original post by oniisanitstoobig)
You didn't use that theorem. You took a different approach.
(Original post by Farhan.Hanif93)
In case it isn't clear, your answer in fact had nothing to do with circles at all. Instead, merely that the distances of the endpoints of a bisected line are equidistant from each point on the perpendicular bisector. Which isn't news or anything like that, and isn't necessarily an unusual way to find the equation of a perpendicular bisector.
Two opposing views, discuss.
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3 years ago
#33
(Original post by 04MR17)
Two opposing views, discuss.
Our views are congruent.
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#34
(Original post by oniisanitstoobig)
You didn't use that fact. You took a different approach.
The fact you used is that the distance from the centre to the circumference is the radius of the circle, for all points.
But whilst I may have used the equation of the circle, I didn't actually use the radius.
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#35
(Original post by oniisanitstoobig)
Our views are congruent.
No, I said that I used the perpendicular bisector theorem, and you said I didn't because I used a different fact. Farhan.Hanif93 said that I used the perpendicular bisector theorem, and you agree with him?
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#36
(Original post by 04MR17)
No, I said that I used the perpendicular bisector theorem, and you said I didn't because I used a different fact. Farhan.Hanif93 said that I used the perpendicular bisector theorem, and you agree with him?
If you think about it, I probably used both. Because in this case, the bisector was a diameter and the endpoints were on the circle. So it did have something to do with a circle, but it didn't have to. The circle was most likely for part b.) and c.).
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