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Question Number 71184 by ajfour last updated on 12/Oct/19

Commented by ajfour last updated on 12/Oct/19

$E a c h s e m i c i r c l e h a v e r a d i u s$ $u n i t y . F i n d m a x i m u m v a l u e o f x .$
Commented by ajfour last updated on 12/Oct/19

Commented by mr W last updated on 13/Oct/19

$(1 - r) \sqrt{1 - {(\frac{b - 1 + 1 - r}{1 - r})}^{2}} = 1 - (1 - r) \sqrt{1 - {(\frac{1 - 1 + r}{1 - r})}^{2}}$ $l e t μ = 1 - b, λ = 1 - r$ $λ \sqrt{1 - {(\frac{- μ + λ}{λ})}^{2}} = 1 - λ \sqrt{1 - {(\frac{1 - λ}{λ})}^{2}}$ $\sqrt{2 μ λ - μ^{2}} = 1 - \sqrt{2 λ - 1}$ $2 λ (1 - μ) + μ^{2} = 2 \sqrt{2 λ - 1}$ $4 {(1 - μ)}^{2} λ^{2} - 4 (2 - μ^{2} + μ^{3}) λ + μ^{4} + 4 = 0$ $λ = \frac{(2 - μ^{2} + μ^{3}) - \sqrt{{(2 - μ^{2} + μ^{3})}^{2} - {(1 - μ)}^{2} (4 + μ^{4})}}{2 {(1 - μ)}^{2}}$ $\Rightarrow λ_{m i n} = 0.5858 a t μ = 0.5858$ $\Rightarrow r_{m a x} = 1 - 0.5858 = 0.4142 = \sqrt{2} - 1$
Commented by ajfour last updated on 13/Oct/19
$x_C =(1−r)sin β=1−(1−r)cos α y_C =b−(1−r)cos β=(1−r)sin α=r (1−r)[(√(1−(((b−r)/(1−r)))^2 ))+(√(1−((r/(1−r)))^2 ))=1 ........(i) (dr/db)=0 ⇒ (1/(2(√(1−(((b−r)/(1−r)))^2 )))){((2(b−r)(1−r)^2 −2(1−r)(b−r)^2 )/((1−r)^4 ))} =(1/(1−r)) ⇒ (((1−b)(b−r))/((1−r)^2 ))=(√(1−(((b−r)/(1−r)))^2 )) ..(ii) let b−r=λ(1−r) ⇒ 1−b=1−r−λ(1−r) from (ii) ⇒ λ(1−λ)=(√(1−λ^2 )) ⇒ λ^2 (1−λ)= 1+λ or λ^3 −λ^2 +λ+1=0 .....$
$x_{C} = (1 - r) \sin β = 1 - (1 - r) \cos α$ $y_{C} = b - (1 - r) \cos β = (1 - r) \sin α = r$ $(1 - r) [\sqrt{1 - {(\frac{b - r}{1 - r})}^{2}} + \sqrt{1 - {(\frac{r}{1 - r})}^{2}} = 1$ $. . . . . . . . (i)$ $\frac{d r}{d b} = 0 \Rightarrow$ $\frac{1}{2 \sqrt{1 - {(\frac{b - r}{1 - r})}^{2}}} {\frac{2 (b - r) {(1 - r)}^{2} - 2 (1 - r) {(b - r)}^{2}}{{(1 - r)}^{4}}}$ $= \frac{1}{1 - r}$ $\Rightarrow \frac{(1 - b) (b - r)}{{(1 - r)}^{2}} = \sqrt{1 - {(\frac{b - r}{1 - r})}^{2}} . . (i i)$ $l e t b - r = λ (1 - r)$ $\Rightarrow 1 - b = 1 - r - λ (1 - r)$ $f r o m (i i)$ $\Rightarrow λ (1 - λ) = \sqrt{1 - λ^{2}}$ $\Rightarrow λ^{2} (1 - λ) = 1 + λ$ $o r λ^{3} - λ^{2} + λ + 1 = 0$ $. . . . .$
	Answered by mr W last updated on 13/Oct/19

	$c e n t e r o f s e m i c i r c l e 1 : (0, 0)$ $c e n t e r o f s e m i c i r c l e 2 : (1, k)$ $c e n t e r o f s m a l l c i r c l e : (h, r)$ $h^{2} + r^{2} = {(1 - r)}^{2} . . . (i)$ ${(1 - h)}^{2} + {(k - r)}^{2} = {(1 - r)}^{2} . . . (i i)$ $(i) - (i i) :$ $2 h - 1 + k (2 r - k) = 0$ $\Rightarrow h = \frac{1 + k^{2}}{2} - k r$ $p u t i n t o (i) :$ ${(\frac{1 + k^{2}}{2} - k r)}^{2} = 1 - 2 r$ $k^{2} r^{2} - (k + k^{3} - 2) r - 1 + \frac{{(1 + k^{2})}^{2}}{4} = 0$ $\Rightarrow r = \frac{k (1 + k^{2}) - 2 + 2 \sqrt{(1 - k) (1 + k^{2})}}{2 k^{2}}$ $\frac{d r}{d k} = 0 \Rightarrow c o m p l i c a t e d e q n . f o r k!$ $n u m e r i c a l l y :$ $r_{m a x} = 0.4142 (= \sqrt{2} - 1) a t k = 0.4142$
	Commented by mr W last updated on 13/Oct/19

	Commented by ajfour last updated on 13/Oct/19

	$T h a n k s s o m u c h S i r, i y e t t h i n k$ $i t s p o s s i b l e t o g e t a n e n t i r e$ $s o l u t i o n, a m t r y i n g . .$
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