Can we say that A line is a circle whose radius is ∞ Or A circle with ∞ radius is a line ?


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Question Number 3695 by Rasheed Soomro last updated on 19/Dec/15

$C a n w e s a y t h a t$ $A l i n e i s a c i r c l e w h o s e r a d i u s i s \infty$ $O r$ $A c i r c l e w i t h \infty r a d i u s i s a l i n e ?$
Commented by Filup last updated on 19/Dec/15

$A line \vec{l} = \infty is like a vector$ $\vec{l} moves in one direction (1 D) .$ $\vec{l} \in R^{1}$ $A circle with radus \vec{r} = \infty is similar .$ $Except a circle will eventually create$ $tangents facing every possible direction$ $in what we can call a 2 D plane .$ $\vec{r} \in R^{2}$ $This is just what i think . I could be$ $wrong .$
Commented by Filup last updated on 19/Dec/15

$A ctually, a c o r r e c t i o n i w a n t t o p o i n t o u t .$ $\vec{r} \notin R^{2}$ $t h e circle \vec{c} \in R^{2} .$ $The radius is 1 D . \vec{r} \in R^{1}$ $A line only has one feature :$ $its length \vec{l} \in R^{1}$ $A circle (\vec{c}) has two features .$ $The radius and the circumference .$ $\vec{r} \in R^{1} \vec{c} \in R^{2}$ $The circle itself lies within a 2 D plane .$ $The radius is in a 1 D plane$
Commented by RasheedSindhi last updated on 19/Dec/15

$P e r h a p s I {c a n}^{'} t c l e a r w h a t I m e a n .$ $I f a n o b j e c t r e v o l v e s o n a c i r c u l a r$ $p a t h, t h e o b j e c t i s s a i d t o b e$ $c o n t i n e o u s l y c h a n g i n g d i r e c t i o n .$ $C o n s i d e r m y^{'} c h a n g e o f {d i r e c t i o n}^{'}$ $s o m e w h a t l i k e t h a t .$ $C o n s i d e r a c i r c l e p a t h o f m o v i n g$ $p o i n t w h i c h k e e p s c h a n g i n g$ $d i r e c t i o n a t e v e r y p o s i t i o n .$
Commented by prakash jain last updated on 19/Dec/15

$Infinite radius means r \to \infty .$ $What is the area of line ? If circle is$ $treated as line .$
Commented by Rasheed Soomro last updated on 19/Dec/15

$A r e a o f l i n e (w h i c h i s c o n s i d e r e d a s c i r c l e) i s$ $half plane . H a l f p l a n e i n w h i c h c e n t r e o c c u r s$ $m a y b e c o n s i d e r e d a s interior of the^{'} {circle}^{'}$ $a n d t h e o t h e r h a l f p l a n e i s exterior a n d l e n g t h$ $o f l i n e (o f c o u r s e \infty) a s perimeter .$ $W h e n I c o n s i d e r a c i r c l e o f \infty r a d i u s I s e e a l i n e!$ $I t s c h a n g e o f d i r e c t i o n i s 0 . A s r g e t s g r e a t e r, c h a n g e$ $o f d i r e c t i o n o f c i r c l e g e t s s m a l l e r a n d s m a l l e r .$
Commented by prakash jain last updated on 19/Dec/15

$If by change of direction you mean angle$ $toward the center = \frac{arc length}{radius}$ $So total = \frac{L}{r} = 2 π and it is independent of radius$ $r \to \infty, Total change of direction is still 2 π$
Commented by RasheedSindhi last updated on 19/Dec/15

$B y c h a n g e o f d i r e c t i o n I m e a n$ $c h a n g e o f a n g l e o f t a n g e n t s .$ $L e t t h e c e n t e r o f t h e c i r c l e i s$ $o r i g i n o f t h e p l a n e a n d t h e p o i n t$ $o f x - a x i s t h r o u g h w h i c h t h e$ $c i r c l e p a s s e s i s p o i n t A .$ $A t A t h e t a n g e n t i s p e r p e n d i c u l a r$ $t o t h e x - a x i s . L e t B i s a n o t h e r$ $p o i n t o f c i r c l e . T a n g e n t o n B$ $w i l l m a k e a n g l e (s a y) θ w i t h$ $x - a x i s . I m e a n b y^{'} c h a n g e o f$ ${d i r e c t i o n}^{'} w a s k e e p i n g c h a n g e o f t h e$ $v a l u e o f θ .$
Commented by prakash jain last updated on 19/Dec/15

$Let AB be a arc of circle and \frac{length of arc AB}{r} = θ$ $then Angle between tangent at A and tangent at$ $B = θ .$ $So change in angle of tangent is same as the angle$ $subtended by arc at the center .$
Commented by prakash jain last updated on 19/Dec/15

$I might have been wrong in my earlier arguments .$ $But let us we have have circle centered at$ $origin (0, 0)$ $x^{2} + y^{2} = r^{2} \Rightarrow \frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = 1$ $\lim_{r \to \infty} what does the above equation become ?$ $Now let us say center of the circle is at (r, 0)$ ${(x - r)}^{2} + y^{2} = r^{2}$ $x^{2} - 2 x r + y^{2} = 0$ $\frac{x^{2} + y^{2}}{r} - 2 x = 0$ $as r \to \infty$ $equation reduced to x = 0 (y - a x i s)$ $So you can s a y t h a t t h e l i n e y - a x i s is a circle$ $w i t h c e n t e r a t (\infty, 0) .$
Commented by Rasheed Soomro last updated on 19/Dec/15

$T h a t i s a p o i n t t o t h i n k!$
Commented by prakash jain last updated on 19/Dec/15

$Also with x = 0 being the circle also makes$ $your earlier argument about half plane$ $being the area valid .$ $And all tangents are parallel .$ $You were correct!$
Commented by Rasheed Soomro last updated on 20/Dec/15

$A n o t h e r t e r m i n w h i c h w e c a n c l e a r$ $o u r i d e a i s^{'} {c u r v e n e s s}^{'}$ $A s r a d i u s o f c i r c l e g e t s g r e a t e r$ $t h e c u r v e n e s s g e t s s m a l l e r .$ $F o r r = 1, f o r c u r v e o f u n i t l e n g t h$ $t h e c u r v e n e s s i s o b v i o u s . B u t f o r$ $r = 1000, i t i s n o t s o o b v i o u s .$ $S o w e c a n s a y :$ $r \to \infty \Rightarrow c u r v e n e s s \to 0$ $o r c u r v e n e s s \to s t r a i g h t n e s s$ $o r$ $c i r c l e \to l i n e$
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