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Question Number 52663 by ajfour last updated on 11/Jan/19

Commented by ajfour last updated on 11/Jan/19

$F i n d r a d i u s R i n t e r m s o f a a n d b .$ $A l s o f i n d t h e c e n t r a l a r e a i n y e l l o w$ $i n t e r m s o f a a n d b . (s o u r c e : a j f o u r)$
	Answered by mr W last updated on 11/Jan/19

	$(a^{2} - r^{2}) + (b^{2} - r^{2}) = {(\sqrt{a^{2} - r^{2}} - r + \sqrt{b^{2} - r^{2}} - r)}^{2}$ $a^{2} + b^{2} - 2 r^{2} = {(\sqrt{a^{2} - r^{2}} + \sqrt{b^{2} - r^{2}} - 2 r)}^{2}$ $a^{2} + b^{2} - 2 r^{2} = a^{2} - r^{2} + b^{2} - r^{2} + 4 r^{2} + 2 [\sqrt{(a^{2} - r^{2}) (b^{2} - r^{2})} - 2 r (\sqrt{a^{2} - r^{2}} + \sqrt{b^{2} - r^{2}})]$ $\Rightarrow \sqrt{(a^{2} - r^{2}) (b^{2} - r^{2})} = 2 r (\sqrt{a^{2} - r^{2}} + \sqrt{b^{2} - r^{2}} - r)$ $e x a m p l e :$ $a = 3, b = 4 \Rightarrow r = 0.9569$
	Commented by ajfour last updated on 11/Jan/19

	$c a n t w e h a v e R i n t e r m s o f a a n d b$ $S i r ? i^{'} v e p r o c e e d e d t o s o m e l e n g t h,$ $s h a l l p o s t s o o n, t h a n k s a f t e r a l l .$
	Answered by ajfour last updated on 18/Jan/19
	$((√(a^2 −r^2 ))−r+(√(b^2 −r^2 ))−r)^2 =a^2 +b^2 −2r^2 ⇒ a^2 −r^2 +b^2 −r^2 +4r^2 −4r((√(a^2 −r^2 ))+(√(b^2 −r^2 )))+2(√(a^2 −r^2 ))(√(b^2 −r^2 )) = a^2 +b^2 −2r^2 ⇒ (√(a^2 −r^2 ))(√(b^2 −r^2 )) +2r^2 =2r((√(a^2 −r^2 ))+(√(b^2 −r^2 ))) squaring a^2 b^2 −(a^2 +b^2 )r^2 +5r^4 +4r^2 (√(a^2 −r^2 ))(√(b^2 −r^2 )) = 4r^2 (a^2 +b^2 −2r^2 +2(√(a^2 −r^2 ))(√(b^2 −r^2 ))) ⇒ a^2 b^2 +13r^4 − 5(a^2 +b^2 )r^2 = 4r^2 (√(a^2 −r^2 ))(√(b^2 −r^2 )) squaring again a^4 b^4 +169r^8 +25(a^2 +b^2 )^2 r^4 +26a^2 b^2 r^4 −10a^2 b^2 (a^2 +b^2 )r^2 −130(a^2 +b^2 )r^6 = 16a^2 b^2 r^4 −16(a^2 +b^2 )r^6 +16r^8 ⇒ 153r^8 −114(a^2 +b^2 )r^6 +5{(a^2 +b^2 )^2 +2a^2 b^2 }r^4 −10a^2 b^2 (a^2 +b^2 )r^2 +a^4 b^4 = 0 let r^2 = t, ⇒ ((153t^2 )/(a^2 b^2 ))+((a^2 b^2 )/t^2 )+5{(((a^2 +b^2 )^2 )/(a^2 b^2 ))+2} = 2(a^2 +b^2 ){((57t)/(a^2 b^2 ))+(5/t)} let (t/(ab)) = x , 5{(((a^2 +b^2 )^2 )/(a^2 b^2 ))+2}=c, ((2(a^2 +b^2 ))/(ab))= d ; ⇒ 153x^2 +(1/x^2 ) + c = d(57x+(5/x)) .......$
	${(\sqrt{a^{2} - r^{2}} - r + \sqrt{b^{2} - r^{2}} - r)}^{2} = a^{2} + b^{2} - 2 r^{2}$ $\Rightarrow a^{2} - r^{2} + b^{2} - r^{2} + 4 r^{2}$ $- 4 r (\sqrt{a^{2} - r^{2}} + \sqrt{b^{2} - r^{2}}) + 2 \sqrt{a^{2} - r^{2}} \sqrt{b^{2} - r^{2}}$ $= a^{2} + b^{2} - 2 r^{2}$ $\Rightarrow \sqrt{a^{2} - r^{2}} \sqrt{b^{2} - r^{2}} + 2 r^{2} = 2 r (\sqrt{a^{2} - r^{2}} + \sqrt{b^{2} - r^{2}})$ $s q u a r i n g$ $a^{2} b^{2} - (a^{2} + b^{2}) r^{2} + 5 r^{4} + 4 r^{2} \sqrt{a^{2} - r^{2}} \sqrt{b^{2} - r^{2}}$ $= 4 r^{2} (a^{2} + b^{2} - 2 r^{2} + 2 \sqrt{a^{2} - r^{2}} \sqrt{b^{2} - r^{2}})$ $\Rightarrow a^{2} b^{2} + 13 r^{4} - 5 (a^{2} + b^{2}) r^{2}$ $= 4 r^{2} \sqrt{a^{2} - r^{2}} \sqrt{b^{2} - r^{2}}$ $s q u a r i n g a g a i n$ $a^{4} b^{4} + 169 r^{8} + 25 {(a^{2} + b^{2})}^{2} r^{4} + 26 a^{2} b^{2} r^{4}$ $- 10 a^{2} b^{2} (a^{2} + b^{2}) r^{2} - 130 (a^{2} + b^{2}) r^{6}$ $= 16 a^{2} b^{2} r^{4} - 16 (a^{2} + b^{2}) r^{6} + 16 r^{8}$ $\Rightarrow$ $153 r^{8} - 114 (a^{2} + b^{2}) r^{6}$ $+ 5 {{(a^{2} + b^{2})}^{2} + 2 a^{2} b^{2}} r^{4}$ $- 10 a^{2} b^{2} (a^{2} + b^{2}) r^{2} + a^{4} b^{4} = 0$ $l e t r^{2} = t, \Rightarrow$ $\frac{153 t^{2}}{a^{2} b^{2}} + \frac{a^{2} b^{2}}{t^{2}} + 5 {\frac{{(a^{2} + b^{2})}^{2}}{a^{2} b^{2}} + 2}$ $= 2 (a^{2} + b^{2}) {\frac{57 t}{a^{2} b^{2}} + \frac{5}{t}}$ $l e t \frac{t}{a b} = x, 5 {\frac{{(a^{2} + b^{2})}^{2}}{a^{2} b^{2}} + 2} = c,$ $\frac{2 (a^{2} + b^{2})}{a b} = d; \Rightarrow$ $153 x^{2} + \frac{1}{x^{2}} + c = d (57 x + \frac{5}{x})$ $. . . . . . .$
	Commented by ajfour last updated on 11/Jan/19

	$y e s s i r, n o t u s e f u l e t a l . .$
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