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Question Number 185816 by mr W last updated on 28/Jan/23

Commented by mr W last updated on 28/Jan/23

$i f t h e r a d i u s o f t h e b i g c i r c l e i s 1,$ $f i n d t h e r a d i i o f t h e s m a l l c i r c l e s .$
	Answered by mr W last updated on 28/Jan/23

	Commented by mr W last updated on 28/Jan/23

	$s p e c i a l c a s e$ $R = b + 2 c = 1$ $b + 2 d = R = b + 2 c$ $\Rightarrow d = c$ $\frac{c}{b + c} = \sin 30 ° = \frac{1}{2}$ $\Rightarrow b = c$ $\Rightarrow a = b = c = d = e = \frac{1}{3}$
	Answered by ajfour last updated on 28/Jan/23
	$E[−(1−e)cos θ, (1−e)sin θ] D[(1−d)cos φ, (1−d)sin φ] {1−a−(1−e)cos θ}^2 +(1−e)^2 sin^2 θ =(a+e)^2 ⇒ (1−a)^2 −2(1−a)(1−e)cos θ =(a+e)^2 −(1−e)^2 ⇒ cos θ=((1−a−e−ae)/(1−a−e+ae)) similarly cos φ=((1−c−d−cd)/(1−c−d+cd)) ...$
	$E [- (1 - e) \cos θ, (1 - e) \sin θ]$ $D [(1 - d) \cos ϕ, (1 - d) \sin ϕ]$ ${1 - a - (1 - e) \cos θ}^{2} + {(1 - e)}^{2} \sin^{2} θ$ $= {(a + e)}^{2}$ $\Rightarrow {(1 - a)}^{2} - 2 (1 - a) (1 - e) \cos θ$ $= {(a + e)}^{2} - {(1 - e)}^{2}$ $\Rightarrow \cos θ = \frac{1 - a - e - a e}{1 - a - e + a e}$ $s i m i l a r l y$ $\cos ϕ = \frac{1 - c - d - c d}{1 - c - d + c d}$ $. . .$
	Answered by mr W last updated on 29/Jan/23

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