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Question Number 173025 by mnjuly1970 last updated on 05/Jul/22

	Answered by mr W last updated on 06/Jul/22

	Commented by mr W last updated on 06/Jul/22

	${(r_{1} + r)}^{2} - {(r_{1} - r)}^{2} = {(2 - r_{1})}^{2} - r_{1}^{2}$ ${r r}_{1} = 1 - r_{1}$ $\Rightarrow r_{1} = \frac{1}{1 + r}$ ${C M}^{2} = 1^{2} + r_{2}^{2}$ $O C = 2 - r_{2}$ $M E = \sqrt{1^{2} + r^{2}}$ $C E = r + r_{2}$ ${C M}^{2} = {M E}^{2} + {C E}^{2} - 2 \times M E \times C E \times \frac{r}{M E}$ $1^{2} + r_{2}^{2} = 1^{2} + r^{2} + {(r + r_{2})}^{2} - 2 r \sqrt{r + r_{2}}$ $r + r_{2} = \sqrt{r + r_{2}}$ $r + r_{2} = 1$ $\Rightarrow r_{2} = 1 - r$ $\frac{r_{1}}{r_{2}} = \frac{1}{(1 + r) (1 - r)} = \frac{7}{6}$ $r^{2} = \frac{1}{7}$ $r = \frac{1}{\sqrt{7}}$ $α = 2 \tan^{- 1} \frac{r}{1} = 2 \tan^{- 1} \frac{1}{\sqrt{7}} = \tan^{- 1} \frac{\sqrt{7}}{3} = 41.4 °$
	Commented by Tawa11 last updated on 11/Jul/22

	$Great sir$
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