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Question Number 206079 by mr W last updated on 06/Apr/24

	Answered by A5T last updated on 06/Apr/24

	Commented by A5T last updated on 06/Apr/24

	$x i s n o t d e p e n d e n t o n t h e p o s i t i o n o f A o n t h e$ $d i a m e t e r, o t h e r w i s e, x w o u l d n o t b e u n i q u e l y$ $d e f i n e d .$ $S o, c o n s i d e r t h e c a s e w h e n A i s c o i n c i d e n t w i t h$ $t h e c e n t e r, t h e n l e n g t h x s u b t e n d s a n a n g l e o f$ $60 ° a t t h e c e n t r e, w h i c h m a k e s i t e q u a l t o t h e$ $r a d i u s \Rightarrow x = 8 . T h i s m u s t a l s o b e t r u e f o r a l l$ $p o s i t i o n s o f A .$
	Answered by mr W last updated on 06/Apr/24

	Commented by mr W last updated on 06/Apr/24

	$x i s a c h o r d c o r r e s p o n d i n g t o a$ $c e n t r a l a n g l e o f 60 ° .$ $\Rightarrow x = R = 8$
	Answered by mr W last updated on 06/Apr/24

	Commented by mr W last updated on 06/Apr/24

	$R^{2} = p^{2} + a^{2} + p a$ $R^{2} = q^{2} + a^{2} - q a$ $\Rightarrow p, - q a r e r o o t s o f z^{2} + a z + a^{2} - R^{2} = 0$ $\Rightarrow p - q = - a, - p q = a^{2} - R^{2}$ $x^{2} = p^{2} + q^{2} - p q = {(p - q)}^{2} + p q$ $= {(- a)}^{2} - (a^{2} - R^{2}) = R^{2}$ $\Rightarrow x = R = 8 ✓$
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