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Question Number 39626 by ajfour last updated on 08/Jul/18

Commented by ajfour last updated on 09/Jul/18

$F i n d b i n t e r m s o f a a n d R .$ $W h a t i s a i n t e r m s o f R i f a + b = R .$
	Answered by ajfour last updated on 09/Jul/18

	$e q . o f A B C : - \frac{x}{a} + \frac{y}{b} = 1$ $\Rightarrow y = b (1 + \frac{x}{a})$ $e q . o f c i r c l e : x^{2} + y^{2} = R^{2}$ $L e t C \equiv (h, k)$ $h^{2} + b^{2} {(1 + \frac{h}{a})}^{2} = R^{2} . . . . (i)$ $E (\frac{R}{2}, \frac{b}{2}) l i e s i n l i n e C E F$ $e q . o f C E F$ $y = (\frac{\frac{b}{2} + R}{R / 2}) x - R$ $C l i e s o n b o t h l i n e s, h e n c e$ $b (1 + \frac{h}{a}) = (\frac{b}{R} + 2) h - R$ $h (\frac{b}{R} - \frac{b}{a} + 2) = b + R$ $h = \frac{b + R}{(\frac{b}{R} - \frac{b}{a} + 2)}$ $l e t \frac{b}{a} = z a n d \frac{R}{a} = p$ $t h e n h = \frac{a (z + p)}{\frac{z}{p} - z + 2} = \frac{a (p z + p^{2})}{(1 - p) z + 2 p}$ $s u b s t i t u t i n g i n (i)$ ${[\frac{a (p z + p^{2})}{(1 - p) z + 2 p}]}^{2} + b^{2} {[1 + \frac{p z + p^{2}}{(1 - p) z + 2 p}]}^{2} = R^{2}$ ${[\frac{(p z + p^{2})}{(1 - p) z + 2 p}]}^{2} + z^{2} {[1 + \frac{p z + p^{2}}{(1 - p) z + 2 p}]}^{2} = p^{2}$ $s o l v i n g a b o v e e q . w e w o u l d g e t$ $z = \frac{b}{a} i n t e r m s o f p = \frac{R}{a}$ $b u t i f a + b = R t h e n a + a z = a p$ $\Rightarrow z = p - 1$ $\Rightarrow \frac{p^{2} {(2 p - 1)}^{2}}{{[2 p - {(1 - p)}^{2}]}^{2}} + {(p - 1)}^{2} {[1 + \frac{p (2 p - 1)}{2 p - {(1 - p)}^{2}}]}^{2} = p^{2}$ $p^{2} {(2 p - 1)}^{2} + {(p - 1)}^{2} {(p^{2} + 3 p - 1)}^{2}$ $= p^{2} {(4 p - 1 - p^{2})}^{2}$ $p^{3} (6 p - p^{2} - 2) (2 - p) = {(p^{3} + 2 p^{2} - 4 p + 1)}^{2}$ $\Rightarrow p^{6} - 8 p^{5} + 14 p^{4} - 4 p^{3} = p^{6} + 4 p^{4} + 16 p^{2}$ $+ 1 + 4 p^{5} - 16 p^{3} - 8 p - 8 p^{4} + 4 p^{2} + 2 p^{3}$ $\Rightarrow 12 p^{5} - 18 p^{4} - 10 p^{3} + 20 p^{2} - 8 p + 1 = 0$ $\Rightarrow p \approx 1.49438$ $a n d \frac{a}{R} = \frac{1}{p} \Rightarrow a \approx 0.669 R .$
	Commented by ajfour last updated on 09/Jul/18

	$M r W 3 S i r, p l e a s e h e l p c h e c k t h i s$ $s o l u t i o n . .$
	Commented by MrW3 last updated on 10/Jul/18

	$I {c o u l d n}^{'} t f i n d a n y m i s t a k e .$
	Commented by ajfour last updated on 10/Jul/18

	$T h a n k y o u s o m u c h S i r; i t i s$ $c o r r e c t t h e n .$
	Answered by MJS last updated on 10/Jul/18

	$R = 1$ $A = (\begin{matrix} - a \\ 0 \end{matrix}) B = (\begin{matrix} 0 \\ b \end{matrix}) D = (\begin{matrix} 1 \\ 0 \end{matrix}) E = (\begin{matrix} \frac{1}{2} \\ \frac{b}{2} \end{matrix}) F = (\begin{matrix} 0 \\ - 1 \end{matrix})$ $line A B$ $l_{1} : y = \frac{b}{a} x + b$ $line E F$ $l_{2} : y = (b + 2) x - 1$ $C = l_{1} \cap l_{2} = (\begin{matrix} \frac{a (b + 1)}{a (b + 2) - b} \\ \frac{(a (b + 2) + 1) b}{a (b + 2) - b} \end{matrix})$ $C on circle x^{2} + y^{2} - 1 = 0$ $(I) \frac{a (b + 1) (a (b^{3} + 3 b^{2} + b - 3) + 2 b (b + 2))}{{(a (b + 2) - b)}^{2}} = 0$ $[a = 0 \lor b = - 1 not of further interest]$ $[a (b + 2) - b \neq 0 \Rightarrow a \neq \frac{b}{b + 2} \Leftrightarrow b \neq \frac{2 a}{1 - a}]$ $this can easily be solved for a$ $a = \frac{2 b (b + 2)}{3 - b - 3 b^{2} - b^{3}} \Rightarrow C = (\begin{matrix} \frac{2 (b + 2)}{b^{2} + 4 b + 5} \\ \frac{(b + 1) (b + 3)}{b^{2} + 4 b + 5} \end{matrix})$ $but for b it leads to$ $b^{3} + \frac{3 a + 2}{a} b^{2} + \frac{a + 4}{a} b - 3 = 0$ $which {can}^{'} t be generally solved$ $with a + b = 1 \Rightarrow b = 1 - a we get$ $(I) \frac{a (a - 2) (a^{4} - 6 a^{3} + 8 a^{2} + 6 a - 6)}{{(a^{2} - 4 a + 1)}^{2}} = 0$ $[a = 0 \lor a = 2 not of further interest]$ $[a^{2} - 4 a + 1 \neq 0 \Rightarrow a \neq 2 - \sqrt{3} \land a \neq 2 + \sqrt{3}]$ $a^{4} - 6 a^{3} + 8 a^{2} + 6 a - 6 = 0$ $again this can be exactly solved but with the$ $usual difficulties, approximately we get$ $a \approx . 669174$ $[a \approx - . 895728 not of further interest]$
	Commented by ajfour last updated on 10/Jul/18

	$T h a n k y o u S i r . I w a s b i t w o r r i e d .$
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