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Question Number 77252 by Maclaurin Stickker last updated on 04/Jan/20

Commented by Maclaurin Stickker last updated on 04/Jan/20

$I {d o n}^{'} t w a n t t h e a n s w e r, I w a n t a t i p$ $o n h o w t o s o l v e .$
Commented by Maclaurin Stickker last updated on 04/Jan/20

$I t r i e d t o f i n d t h e r a d i u s o f t h e$ $l a r g e s t c i r c u m f e r e n c e a s a f u n c t i o n$ $o f t h e r a d i u s o f t h e y e l l o w$ $c i r c u m f e r e n c e b u t t h e r e s u l t s {w e r e n}^{'} t$ $g o o d . . .$
	Answered by mr W last updated on 05/Jan/20

	Commented by Maclaurin Stickker last updated on 05/Jan/20

	$s i r, I u s e d \frac{A}{B} = 4 a n d I g o t \frac{9}{20}$ $i s m y a n s w e r r i g h t ?$
	Commented by mr W last updated on 05/Jan/20

	$O a s o r i g i n a n d O F a s x - a x i s$ $x_{A} = - R + r_{a}$ $x_{C} = 2 r_{a} - R$ $x_{B}^{2} + y_{B}^{2} = {(R - r_{b})}^{2}$ ${(x_{B} + R - r_{a})}^{2} + y_{B}^{2} = {(r_{a} + r_{b})}^{2}$ ${(x_{B} + R - r_{a})}^{2} - x_{B}^{2} = {(r_{a} + r_{b})}^{2} - {(R - r_{b})}^{2}$ $\Rightarrow (2 x_{B} + R - r_{a}) (R - r_{a}) = (R + r_{a}) (r_{a} + 2 r_{b} - R)$ $\Rightarrow x_{B} = \frac{(R + r_{a}) (2 r_{b} - R + r_{a}) - {(R - r_{a})}^{2}}{2 (R - r_{a})}$ $\Rightarrow y_{B} = \sqrt{{(R - r_{b})}^{2} - x_{B}^{2}}$ $C F = 2 (R - r_{a})$ $\frac{E F}{G F} = \frac{2 R}{E F} \Rightarrow {E F}^{2} = 2 R (R - r_{a})$ ${E G}^{2} = {E F}^{2} - {G F}^{2} = 2 R (R - r_{a}) - {(R - r_{a})}^{2}$ ${E G}^{2} = (R - r_{a}) (R + r_{a}) = R^{2} - r_{a}^{2}$ $\Rightarrow E G = \sqrt{R^{2} - r_{a}^{2}}$ $m = \tan ∠ E C G = \frac{E G}{C G} = \frac{\sqrt{R^{2} - r_{a}^{2}}}{R - r_{a}} = \sqrt{\frac{R + r_{a}}{R - r_{a}}}$ $x_{C} = 2 r_{a} - R$ $e q n . o f C E :$ $y = m (x - x_{C})$ $\Rightarrow m x - y - m (2 r_{a} - R) = 0$ $r_{b} = \frac{m (x_{B} + R - 2 r_{a}) - y_{B}}{\sqrt{1 + m^{2}}}$ $r_{b} = \frac{\sqrt{\frac{R - r_{a}}{R + r_{a}}} (x_{B} + R - 2 r_{a}) - y_{B}}{\sqrt{\frac{2 R}{R + r_{a}}}}$ $\Rightarrow r_{b} = \sqrt{\frac{R - r_{a}}{2 R}} (x_{B} + R - 2 r_{a}) - \sqrt{\frac{R + r_{a}}{2 R}} y_{B}$ $l e t α = \frac{r_{a}}{R}, β = \frac{r_{b}}{R}$ $\frac{x_{B}}{R} = μ = \frac{(1 + α) (2 β - 1 + α) - {(1 - α)}^{2}}{2 (1 - α)}$ $\frac{y_{B}}{R} = ν = \sqrt{{(1 - β)}^{2} - μ^{2}}$ $\Rightarrow β = \sqrt{\frac{1 - α}{2}} (μ + 1 - 2 α) - \sqrt{\frac{1 + α}{2}} ν$ $. . . . . .$
	Commented by mr W last updated on 05/Jan/20

	$e x a c t s o l u t i o n s e e m s i m p o s s i b l e .$
	Commented by Maclaurin Stickker last updated on 05/Jan/20

	$i s B C p e r p e n d i c u l a r t o O C ? I t h i n k s o . .$
	Commented by Maclaurin Stickker last updated on 05/Jan/20

	${D o e s n}^{'} t \frac{A}{B} \in {4, 1, \frac{9}{16}} m e a n t h a t$ $\frac{A}{B} = 4, \frac{A}{B} = 1 o r \frac{A}{B} = \frac{9}{16} ?$
	Commented by mr W last updated on 05/Jan/20

	$B C i s n o t p e r p e n d i c u l a r t o O C!$ $y o u c a n s e e t h i s c l e a r l y i f r_{b} = r_{a .}$
	Commented by Maclaurin Stickker last updated on 05/Jan/20

	$I f r_{a} \neq r_{b} B C c o u l d b e p e r p e n d i c u l a r$ $t o O C ?$
	Commented by mr W last updated on 06/Jan/20

	$n o t g e n e r a l l y!$
	Commented by Tawa11 last updated on 29/Dec/21

	$Great sir$
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