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Question Number 51422 by ajfour last updated on 26/Dec/18

Commented by ajfour last updated on 26/Dec/18

$F i n d R t h e r a d i u s o f s e m i c i r c l e$ $i n t e r m s o f r a d i i a a n d b o f t h e$ $c i r c l e s .$
	Answered by ajfour last updated on 26/Dec/18
	$let sin θ = (a/(R−a)) , sin φ = (b/(R−b)) (R−a)cos θ+(R−b)cos φ = (√(4ab)) ⇒ (√(R(R−2a)))+(√(R(R−2b))) = (√(4ab)) squaring 2R^2 −2(a+b)R +2(√(R^2 {R^2 −2(a+b)R+4ab})) = 4ab rearranging and squaring R^2 {R^2 −2(a+b)R+4ab} = {2ab−R^2 +(a+b)R}^2 ⇒ 4abR^2 = 4a^2 b^2 +(a+b)^2 R^2 −4abR^2 +4ab(a+b)R ⇒ (a^2 +b^2 −6ab)R^2 +4ab(a+b)R +4a^2 b^2 = 0 ⇒ R = ((−2ab(a+b)−(√(4a^2 b^2 (8ab))))/(a^2 +b^2 −6ab)) or R = [(√(8ab))+(a+b)](((2ab)/(6ab−a^2 −b^2 ))) .$
	$l e t \sin θ = \frac{a}{R - a}, \sin ϕ = \frac{b}{R - b}$ $(R - a) \cos θ + (R - b) \cos ϕ = \sqrt{4 a b}$ $\Rightarrow \sqrt{R (R - 2 a)} + \sqrt{R (R - 2 b)} = \sqrt{4 a b}$ $s q u a r i n g$ $2 R^{2} - 2 (a + b) R + 2 \sqrt{R^{2} {R^{2} - 2 (a + b) R + 4 a b}} = 4 a b$ $r e a r r a n g i n g a n d s q u a r i n g$ $R^{2} {R^{2} - 2 (a + b) R + 4 a b}$ $= {2 a b - R^{2} + (a + b) R}^{2}$ $\Rightarrow 4 {a b R}^{2} = 4 a^{2} b^{2} + {(a + b)}^{2} R^{2}$ $- 4 {a b R}^{2} + 4 a b (a + b) R$ $\Rightarrow (a^{2} + b^{2} - 6 a b) R^{2} + 4 a b (a + b) R$ $+ 4 a^{2} b^{2} = 0$ $\Rightarrow R = \frac{- 2 a b (a + b) - \sqrt{4 a^{2} b^{2} (8 a b)}}{a^{2} + b^{2} - 6 a b}$ $o r R = [\sqrt{8 a b} + (a + b)] (\frac{2 a b}{6 a b - a^{2} - b^{2}}) .$
	Answered by behi83417@gmail.com last updated on 27/Dec/18

	$\Rightarrow A B = \sqrt{2 (a^{2} + b^{2})}$ $T O = O S = R$ $O A = R - a, O B = R - b$ $s i n α = \frac{a}{R - a}, s i n β = \frac{b}{R - b}$ $c o s α = \frac{\sqrt{{(R - a)}^{2} - a^{2}}}{R - a}, c o s β = \frac{\sqrt{{(R - b)}^{2} - b^{2}}}{R - b}$ $c o s φ = c o s (π - α - β) = - c o s (α + β) =$ $= \frac{a b - \sqrt{R^{2} - 2 R a} . \sqrt{R^{2} - 2 R b}}{(R - a) (R - b)} (i)$ $c o s φ = \frac{{O A}^{2} + {O B}^{2} - {A B}^{2}}{2 O A . O B} = \frac{{(R - a)}^{2} + {(R - b)}^{2} - 2 (a^{2} + b^{2})}{2 (R - a) (R - b)} =$ $= \frac{2 R^{2} - 2 R a - 2 R b - a^{2} - b^{2}}{2 (R - a) (R - b)} (i i)$ $i = i i \Rightarrow$ $2 R^{2} - 2 R a - 2 R b - a^{2} - b^{2} =$ $= a b - \sqrt{R^{2} - 2 R a} . \sqrt{R^{2} - 2 R b}$ $a^{2} + b^{2} + a b + 2 R a + 2 R b - 2 R^{2} = \sqrt{R^{2} - 2 R a} . \sqrt{R^{2} - 2 R b}$ $R^{2} - 2 R a = p, R^{2} - 2 R b = q, a^{2} + b^{2} + a b = t$ $a^{2} + b^{2} + a b - (p + q) = \sqrt{p q}$ $t - (p + q) = \sqrt{p q} \Rightarrow t^{2} - 2 t (p + q) + {(p + q)}^{2} = p q$ $p + q = 2 R^{2} - 2 R a - 2 R b$ ${(p + q)}^{2} = 4 R^{4} + 4 R^{2} a^{2} + 4 R^{2} b^{2} - 8 R^{3} a - 8 R^{3} b + 8 R^{2} a b$ $p q = (R^{2} - 2 R a) (R^{2} - 2 R b) = R^{4} - 2 R^{3} b - 2 R^{3} a + 4 R^{2} a b$ $t^{2} - 4 R (R - a - b) t + 3 R^{4} + 4 R^{2} (a^{2} + b^{2} + a b)$ $- 6 R^{3} a - 6 R^{3} b = 0$ $t^{2} + 4 R (a + b) t + 3 R^{3} (R - 2 a - 2 b) = 0$ $\Rightarrow △^{'} = 4 R^{2} {(a + b)}^{2} - 3 R^{3} (R - 2 a - 2 b) =$ $= R^{2} [4 {(a + b)}^{2} - 3 R^{2} + 6 R (a + b)] = R^{2} m^{2}$ $a^{2} + b^{2} + a b = - 2 R (a + b) + R m$ $m = \frac{a^{2} + b^{2} + a b}{R} + 2 (a + b)$
	Commented by ajfour last updated on 27/Dec/18

	$f o r b = a, w e m u s t g e t R = a (1 + \sqrt{2})$ $y o u r s o l u t i o n g e t s$ $m = \frac{3 a^{2}}{R} + 4 a$ $3 a^{2} = - 4 a R + 3 a^{2} + 4 a R$ $s o w e d o n t o b t a i n R, p l e a s e r e s o l v e$ $S i r .$
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