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Question Number 29627 by ajfour last updated on 10/Feb/18

Commented by ajfour last updated on 10/Feb/18

$A l s o f i n d \frac{a}{R} i f l e n g t h o f A B i s$ $e q u a l t o a .$
	Answered by mrW2 last updated on 10/Feb/18

	$O = c e n t e r o f c i r c l e$ $O P = \sqrt{a^{2} + R^{2}}$ $P T = \sqrt{a^{2} + R^{2}} + R$ $\frac{B T}{P T} = \frac{O Q}{Q T}$ $B T = \frac{R (\sqrt{a^{2} + R^{2}} + R)}{a}$ ${A B}^{2} = {A T}^{2} + {B T}^{2} = 4 R^{2} + \frac{R^{2} (2 R^{2} + a^{2} + 2 R \sqrt{a^{2} + R^{2}})}{a^{2}}$ $= \frac{R^{2} (2 R^{2} + 5 a^{2} + 2 R \sqrt{a^{2} + R^{2}})}{a^{2}}$ $\Rightarrow A B = \frac{R \sqrt{2 R^{2} + 5 a^{2} + 2 R \sqrt{a^{2} + R^{2}}}}{a}$ $i f A B = a$ $\frac{R^{2} (2 R^{2} + 5 a^{2} + 2 R \sqrt{a^{2} + R^{2}})}{a^{2}} = a^{2}$ $\Rightarrow 2 + 5 {(\frac{a}{R})}^{2} + 2 \sqrt{1 + {(\frac{a}{R})}^{2}} = {(\frac{a}{R})}^{4}$ $\Rightarrow \frac{a}{R} \approx 2.4875 (v i a g r a p h)$
	Commented by ajfour last updated on 10/Feb/18

	$T h a n k y o u S i r .$
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