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Question Number 46308 by ajfour last updated on 23/Oct/18

Commented by ajfour last updated on 23/Oct/18

$O i s t h e c e n t r e o f c i r c l e, C t h e$ $c e n t r e o f t h e u n i t s q u a r e .$ $I f (i) α = \tan^{- 1} \frac{4}{3}, (i i) α = 60 °,$ $f i n d r a d i u s R .$
	Answered by MrW3 last updated on 23/Oct/18

	Commented by MrW3 last updated on 23/Oct/18

	$l e t a = 1 a n d b = \frac{a}{2}$ $O Q = R - b$ $C Q = b$ $O C = \sqrt{b^{2} + {(R - b)}^{2}}$ $O T = 2 b - R$ $O P = R$ $P T = \sqrt{R^{2} - {(2 b - R)}^{2}}$ $C Q - P T = b - \sqrt{R^{2} - {(2 b - R)}^{2}}$ ${C P}^{2} = b^{2} + {[b - \sqrt{R^{2} - {(2 b - R)}^{2}}]}^{2}$ $= 2 b [2 R - \sqrt{4 b (R - b)} - b]$ $R^{2} = b^{2} + {(R - b)}^{2} + 2 b [2 R - \sqrt{4 b (R - b)} - b] - 2 \sqrt{2 b [b^{2} + {(R - b)}^{2}] [2 R - \sqrt{4 b (R - b)} - b]} \cos α$ $0 = 2 b (R + b) - 2 b \sqrt{4 b (R - b)} - 2 b^{2} - 2 \sqrt{2 b [b^{2} + {(R - b)}^{2}] [2 R - \sqrt{4 b (R - b)} - b]} \cos α$ $l e t λ = \frac{R}{b} = \frac{2 R}{a} = 2 R$ $λ - 2 \sqrt{λ - 1} - \sqrt{2 [1 + {(λ - 1)}^{2}] [2 λ - 2 \sqrt{λ - 1} - 1]} \cos α = 0$ $w i t h α = 60 °$ $2 λ - 4 \sqrt{λ - 1} - \sqrt{2 [1 + {(λ - 1)}^{2}] [2 λ - 2 \sqrt{λ - 1} - 1]} = 0$ $\Rightarrow λ \approx 1.0636$ $\Rightarrow R \approx 0.5318$ $w i t h α = \tan^{- 1} \frac{4}{3}$ $\Rightarrow λ \approx 1.0196$ $\Rightarrow R \approx 0.5098$
	Commented by ajfour last updated on 23/Oct/18

	$c o r r e c t a n s w e r s i r, t h a n k y o u!$
	Answered by ajfour last updated on 23/Oct/18

	Commented by ajfour last updated on 23/Oct/18

	$α = θ + ϕ$ $\tan θ = \frac{a / 2}{R - \frac{a}{2}}$ $l e t λ = \frac{R}{a}$ $\Rightarrow \tan θ = \frac{1}{2 λ - 1} = \frac{1}{t^{2}} (s a y)$ $\tan ϕ = \frac{P M}{C M} = \frac{\sqrt{R^{2} - {(a - R)}^{2}} - \frac{a}{2}}{a / 2}$ $= 2 \sqrt{2 λ - 1} - 1 = 2 t - 1$ $l e t m = \tan α$ $\Rightarrow m = \tan (θ + ϕ)$ $m - m \tan θ \tan ϕ = \tan θ + \tan ϕ$ $m - \frac{m (2 t - 1)}{t^{2}} = \frac{1}{t^{2}} + 2 t - 1$ $\Rightarrow {m t}^{2} - 2 m t + m = 1 + 2 t^{3} - t^{2}$ $2 t^{3} - (1 + m) t^{2} + 2 m t - (m - 1) = 0$ $f o r α = \tan^{- 1} \frac{4}{3}, m = \frac{4}{3}$ $2 t^{3} - \frac{7 t^{2}}{3} + \frac{8 t}{3} - \frac{1}{3} = 0$ $o r 6 t^{3} - 7 t^{2} + 8 t - 1 = 0$ $t \approx 0.14012$ $λ = \frac{R}{a} = \frac{t^{2} + 1}{2}$ $R \approx 0.5098 .$
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