Question-11221

Question Number 11221 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

Commented by mrW1 last updated on 18/Mar/17

c e n t e r p o i n t o f t h e c i r c l e M (a, a)

r a d i u s o f t h e c i r c l e R = (a - 1) \sqrt{2}

\sin \frac{θ}{2} = \frac{R}{a \sqrt{2}} =∣ \frac{a - 1}{a} ∣⩽ 1

θ = {2 \sin}^{- 1} (∣ \frac{a - 1}{a} ∣), ∣ a ∣⩾ \frac{1}{2}

w h e n ∣ a ∣\to \frac{1}{2}, θ \to 180 °

w h e n ∣ a ∣\to 1, θ \to 0 °

w h e n ∣ a ∣\to \infty, θ \to 180 °

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17

h e l l o m r W 1 . t h a n k y o u s o m u c h . b u t

i t h i n k, t h i s i s a v e r y s p i c i a l c a s e o f

q u i s t i o n . c e n t r e o f c i r c l e m y n o t a l w y e s

l i e o n t h e l i n e y = x .

Commented by mrW1 last updated on 17/Mar/17

t h e c u r v e x y = 1 i s s y m m e t r i c a b o u t t h e

l i n e y = x . f o r a c i r c l e w h i c h t a n g e n t s

t h e c u r v e x y = 1 a t (1, 1) i t s c e n t e r m u s t

b e o n t h e l i n e y = x .

Commented by mrW1 last updated on 18/Mar/17

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