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Question Number 62839 by ajfour last updated on 25/Jun/19

Commented by ajfour last updated on 25/Jun/19
$let θ be angle between b and extended c . (c+bcos θ)^2 +b^2 sin^2 θ=r^2 (c+asin θ)^2 +a^2 cos^2 θ=r^2 ⇒ b^2 +c^2 +2bccos θ=r^2 a^2 +c^2 +2acsin θ=r^2 ⇒ 2c(asin θ−bcos θ)=b^2 −a^2 2c(√(a^2 +b^2 ))sin (θ−tan^(−1) (b/a))=b^2 −a^2 ⇒ θ=α+sin^(−1) (p/h) = α+β .....(I) but on adding 2r^2 = a^2 +b^2 +2c^2 +2c(√(a^2 +b^2 ))sin (θ+tan^(−1) (b/a)) ⇒ r^2 =c^2 +((a^2 +b^2 )/2)+(h/2)sin (2α+β) = c^2 +((a^2 +b^2 )/2)+(h/2){((2ab)/(a^2 +b^2 ))×(√(1−(p^2 /h^2 )))+(((2a^2 )/(a^2 +b^2 ))−1)(p/h)} r^2 =c^2 +((a^2 +b^2 )/2)+((ab(√(h^2 −p^2 )))/(a^2 +b^2 ))+(((a^2 −b^2 )p)/(2(a^2 +b^2 ))) r^2 =c^2 +((a^2 +b^2 )/2)−(((b^2 −a^2 )^2 )/(2(a^2 +b^2 )))+((ab)/((a^2 +b^2 )))(√(4c^2 (a^2 +b^2 )−(b^2 −a^2 )^2 )) ⇒ r^2 =c^2 +((2a^2 b^2 )/(a^2 +b^2 ))+((ab)/((a^2 +b^2 )))(√(4c^2 (a^2 +b^2 )−(b^2 −a^2 )^2 )) r=(√(c^2 +((2a^2 b^2 )/(a^2 +b^2 ))+((ab)/((a^2 +b^2 )))(√(4c^2 (a^2 +b^2 )−(b^2 −a^2 )^2 )))) .$
$l e t θ b e a n g l e b e t w e e n b a n d$ $e x t e n d e d c .$ ${(c + b \cos θ)}^{2} + b^{2} \sin^{2} θ = r^{2}$ ${(c + a \sin θ)}^{2} + a^{2} \cos^{2} θ = r^{2}$ $\Rightarrow$ $b^{2} + c^{2} + 2 b c \cos θ = r^{2}$ $a^{2} + c^{2} + 2 a c \sin θ = r^{2}$ $\Rightarrow 2 c (a \sin θ - b \cos θ) = b^{2} - a^{2}$ $2 c \sqrt{a^{2} + b^{2}} \sin (θ - \tan^{- 1} \frac{b}{a}) = b^{2} - a^{2}$ $\Rightarrow θ = α + \sin^{- 1} \frac{p}{h} = α + β . . . . . (I)$ $b u t o n a d d i n g$ $2 r^{2} = a^{2} + b^{2} + 2 c^{2} + 2 c \sqrt{a^{2} + b^{2}} \sin (θ + \tan^{- 1} \frac{b}{a})$ $\Rightarrow r^{2} = c^{2} + \frac{a^{2} + b^{2}}{2} + \frac{h}{2} \sin (2 α + β)$ $= c^{2} + \frac{a^{2} + b^{2}}{2} + \frac{h}{2} {\frac{2 a b}{a^{2} + b^{2}} \times \sqrt{1 - \frac{p^{2}}{h^{2}}} + (\frac{2 a^{2}}{a^{2} + b^{2}} - 1) \frac{p}{h}}$ $r^{2} = c^{2} + \frac{a^{2} + b^{2}}{2} + \frac{a b \sqrt{h^{2} - p^{2}}}{a^{2} + b^{2}} + \frac{(a^{2} - b^{2}) p}{2 (a^{2} + b^{2})}$ $r^{2} = c^{2} + \frac{a^{2} + b^{2}}{2} - \frac{{(b^{2} - a^{2})}^{2}}{2 (a^{2} + b^{2})} + \frac{a b}{(a^{2} + b^{2})} \sqrt{4 c^{2} (a^{2} + b^{2}) - {(b^{2} - a^{2})}^{2}}$ $\Rightarrow r^{2} = c^{2} + \frac{2 a^{2} b^{2}}{a^{2} + b^{2}} + \frac{a b}{(a^{2} + b^{2})} \sqrt{4 c^{2} (a^{2} + b^{2}) - {(b^{2} - a^{2})}^{2}}$ $r = \sqrt{c^{2} + \frac{2 a^{2} b^{2}}{a^{2} + b^{2}} + \frac{a b}{(a^{2} + b^{2})} \sqrt{4 c^{2} (a^{2} + b^{2}) - {(b^{2} - a^{2})}^{2}}} .$
Commented by ajfour last updated on 25/Jun/19

$F i n d r a d i u s o f c i r c l e i n t e r m s o f$ $a, b, c . C i s c e n t e r o f c i r c l e, a n d$ $s e g m e n t s o f l e n g t h s a a n d b a r e$ $p e r p e n d i c u l a r t o e a c h o t h e r .$
Commented by behi83417@gmail.com last updated on 25/Jun/19

$great question, sir Ajfour!$ $waitng for solution (s) . . . . . . . . . .$
	Answered by mr W last updated on 25/Jun/19

	$r = r a d i u s o f c i r c l e$ $α = a n g l e b e t w e e n a a n d c$ $β = a n g l e b e t w e e n b a n d c$ $α + β + \frac{π}{2} = 2 π$ $\Rightarrow α = 2 π - (\frac{π}{2} + β)$ $\Rightarrow \cos α = - \sin β$ $\Rightarrow \cos^{2} α + \cos^{2} β = 1$ $\cos α = \frac{a^{2} + c^{2} - r^{2}}{2 a c}$ $\cos β = \frac{b^{2} + c^{2} - r^{2}}{2 b c}$ $\Rightarrow {(\frac{a^{2} + c^{2} - r^{2}}{2 a c})}^{2} + {(\frac{b^{2} + c^{2} - r^{2}}{2 b c})}^{2} = 1$ $b^{2} [a^{4} + 2 a^{2} c^{2} + c^{4} - 2 (a^{2} + c^{2}) r^{2} + r^{4}] + a^{2} [b^{4} + 2 b^{2} c^{2} + c^{4} - 2 (b^{2} + c^{2}) r^{2} + r^{4}] = 4 a^{2} b^{2} c^{2}$ $(a^{2} + b^{2}) r^{4} - 2 [2 a^{2} b^{2} + (a^{2} + b^{2}) c^{2}] r^{2} + (a^{2} + b^{2}) (a^{2} b^{2} + c^{4}) = 0$ $\Rightarrow r^{2} = \frac{2 a^{2} b^{2} + (a^{2} + b^{2}) c^{2} + \sqrt{{[2 a^{2} b^{2} + (a^{2} + b^{2}) c^{2}]}^{2} - {(a^{2} + b^{2})}^{2} (a^{2} b^{2} + c^{4})}}{a^{2} + b^{2}}$ $\Rightarrow r^{2} = \frac{2 a^{2} b^{2} + (a^{2} + b^{2}) c^{2} + a b \sqrt{4 (a^{2} + b^{2}) c^{2} - {(a^{2} - b^{2})}^{2}}}{a^{2} + b^{2}}$ $\Rightarrow r = \sqrt{\frac{2 a^{2} b^{2} + (a^{2} + b^{2}) c^{2} + a b \sqrt{4 (a^{2} + b^{2}) c^{2} - {(a^{2} - b^{2})}^{2}}}{a^{2} + b^{2}}}$
	Commented by ajfour last updated on 25/Jun/19

	$T o o g o o d S i r . T h a n k s f o r c o n f i r m i n g .$
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