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

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