Question-197916

Question Number 197916 by sonukgindia last updated on 04/Oct/23

Answered by MM42 last updated on 04/Oct/23

Commented by MM42 last updated on 05/Oct/23

l e t O B = r & C D = x

9 = {(2 + 2 r + x)}^{2} + 1 \Rightarrow 2 r + x = \sqrt{8} - 2 \Rightarrow 2 r = \sqrt{8} - 2 - x

{(1 + r)}^{2} = {(x + r)}^{2} + 1 \Rightarrow 2 r = 2 r x + x^{2}

\sqrt{8} - 2 - x = \sqrt{8} x - 2 x - x^{2} + x^{2}

(\sqrt{8} - 1) x = \sqrt{8} - 2 \Rightarrow x = \frac{\sqrt{8} - 2}{\sqrt{8} - 1}

\Rightarrow r = \frac{12 - 4 \sqrt{8}}{\sqrt{8} - 1} ✓

Answered by mr W last updated on 05/Oct/23

g e n e r a l l y

{(\frac{1}{R} + \frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}})}^{2} = 2 (\frac{1}{R^{2}} + \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}})

{(\frac{1}{R} + \frac{1}{1} + \frac{1}{1} + \frac{1}{2})}^{2} = 2 (\frac{1}{R^{2}} + \frac{1}{1^{2}} + \frac{1}{1^{2}} + \frac{1}{2^{2}})

\frac{1}{R^{2}} - \frac{5}{R} - \frac{7}{4} = 0

\frac{1}{R} = \frac{5 \pm 4 \sqrt{2}}{2}

\Rightarrow R = \frac{2}{5 \pm 4 \sqrt{2}} = \frac{8 \sqrt{2} - 10}{7}, - \frac{8 \sqrt{2} + 10}{7}

r a d i u s o f s m a l l c i r c l e = \frac{8 \sqrt{2} - 10}{7} \approx 0.188

r a d i u s o f b i g c i r c l e = \frac{8 \sqrt{2} + 10}{7} \approx 3.044

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