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Question Number 200092 by ajfour last updated on 13/Nov/23

Commented by ajfour last updated on 13/Nov/23

$I f s e m i c i r c l e h a s r a d i u s 1, f i n d$ $r a d i i o f b l u e a n d g r e e n c i r c l e s .$ $O n e e n d o f e a c h s e m i c i r c l e i s a t$ $c e n t e r o f o t h e r .$
	Answered by mr W last updated on 13/Nov/23

	Commented by mr W last updated on 13/Nov/23

	$\sqrt{{(1 + a)}^{2} - a^{2}} = 1 + \sqrt{{(1 - a)}^{2} - a^{2}}$ $\sqrt{1 + 2 a} = 1 + \sqrt{1 - 2 a}$ $4 a - 1 = 2 \sqrt{1 - 2 a}$ $16 a^{2} = 3$ $\Rightarrow a = \frac{\sqrt{3}}{4} ✓$ $\sqrt{{(1 - b)}^{2} - b^{2}} = \frac{1}{2}$ $1 - 2 b = \frac{1}{4}$ $\Rightarrow b = \frac{3}{8} ✓$
	Commented by ajfour last updated on 13/Nov/23

	$e x c e l l e n t w o r k i n g s! t h a n k y o u s i r .$
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