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Circle and Infinity? 
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#1
Oct407, 03:26 PM

P: 2

Why is it that a circle with an infinitly large radius is a line?



#2
Aug2209, 04:55 PM

P: 39

Why is it that this planet looks flat to our point of perception? 


#3
Aug2309, 06:52 AM

P: 607

Depends on your definition of "circle".



#4
Aug2309, 12:06 PM

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Circle and Infinity?
There is no such thing as a "circle with infinite radius". What is true is that the "curvature" of a circle with radius R is 1/R. In the limit, as the radius goes to infinity (which NOT the same as saying the radius is infinite!), the curvature goes to 0 and the only "curve" having curvature 0 at every point is a straight line.
By the way, in hyperbolic geometry, the limit of a sequence of circles, having the same center but increasing radius, is NOT a straight line but a "horocycle". 


#5
Aug2409, 08:08 AM

P: 607

There is no such thing as a "circle with infinite radius".
Depends on your definition of "circle". In complex analysis we sometimes see a definition of "circle" that includes lines. That way it doesn't matter whether we think of the complex numbers as a plane (Argand) or a sphere (Riemann). 


#6
Aug2409, 08:21 AM

P: 164

A curve is line because as the radius becomes larger, it becomes like a line. Try it in your compass.



#7
Aug3009, 10:18 AM

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#8
Aug3009, 12:41 PM

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I will point out again that in hyperbolic geometry, if you take circles passing through a given point with larger and larger radius, they "look" more and more like a line but the limiting figure is a "horocycle", NOT a line. 


#9
Sep1011, 08:10 PM

P: 5

Consider a circumscribed nagon. Join each vertex of one of the sides AB to the centre O. The length of the sides of AO and BO would be the radius of the circle r.
The area of the triangle AOB using the trigonometry equation would be ½ AO*BO sin {360/n} or ½ r^2 sin{360/n} The area of the nagon would be n*1/2 r^2 sin {360/n} As n> infinity the area of the ngon approaches the area of the circle Or n/2 * r^2 sin {360/n} > pi r^2 We can eliminate r^2 from the equation thus: n/2 * sin{ 360/n} > pi as n> infinity Let n = 1000000 which is very big but still far from infinity. n/2* sin{ 360/n} –pi =0.0000000000206932385 which is close. This result points to the fact that as the number of straight sides of the nagon approaches infinity the circle and nagon become closer to being identical. Consider then the perimeter of the nagon and the circumference of the circumscribing circle. Let the base of our triangle be s and the two radius long sides r. s/sin(360/n)= r/sin ((180360)/2) this simplifies to s=r*sin(360/n)/((sin180*n 360)/2)however as N>infinity sin180*n360)/2>sin90degrees=1 so n*s =n/2*sin(360/n) and this we know >pi as n>infinity. We might consider the circle as being an infinitysided polygon. 


#10
Sep1111, 07:02 AM

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Did you notice that this thread is three years old?
How do people find such old threads? R.I.P. 


#11
Sep1111, 07:07 PM

P: 5

When I was 12 in a maths lesson I was doing simple equations the answers of which were of the form x=3 etc. We were required to double underline our answers. Being 12 I used the curved edge of my protractor and for the short answers the arc of 10 degrees seemed straight to me. I figured if I had a bigger protractor I could draw a 10 degree arc that would be longer and still look straight my mind followed bigger and bigger protractors to the conclusion that a straight line is the arc of a circle of infinite radius. I turned to the back of my maths book and wrote this down. Just then I received a blinding cuff to the back of my head by my teacher and was toldto , "Get on with your maths boy".
As I grew oder this still perplexed me and I thought about it often. What I wrote in my initial posting was part of my musing. How did I find this particular posting? I entered my statement made all those years ago in Google and it sent me here. 


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