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Question Number 48611 by Tawa1 last updated on 25/Nov/18

	Answered by mr W last updated on 26/Nov/18

	$r_{1}, r_{2} = r a d i i o f t h e t w o b i g g e r c i r c l e s$ $r_{3} = r a d i u s o f t h e s m a l l e s t c i r c l e$ $\sqrt{{(r_{1} + r_{3})}^{2} - {(r_{1} - r_{3})}^{2}} + \sqrt{{(r_{2} + r_{3})}^{2} - {(r_{2} - r_{3})}^{2}} = \sqrt{{(r_{1} + r_{2})}^{2} + {(r_{1} - r_{2})}^{2}}$ $\sqrt{2 r_{1} 2 r_{3}} + \sqrt{2 r_{2} 2 r_{3}} = \sqrt{2 r_{1} 2 r_{2}}$ $\sqrt{r_{1} r_{3}} + \sqrt{r_{2} r_{3}} = \sqrt{r_{1} r_{2}}$ $\Rightarrow \frac{1}{\sqrt{r_{1}}} + \frac{1}{\sqrt{r_{2}}} = \frac{1}{\sqrt{r_{3}}}$ $o r$ $\Rightarrow r_{3} = \frac{r_{1} r_{2}}{r_{1} + r_{2} + 2 \sqrt{r_{1} r_{2}}}$ $\Rightarrow r_{3} = \frac{1 \times 2}{1 + 2 + 2 \sqrt{1 \times 2}} = \frac{2}{3 + 2 \sqrt{2}} = 2 (3 - 2 \sqrt{2}) \approx 0.343$
	Commented by Tawa1 last updated on 26/Nov/18

	$God bless you sir$
	Commented by Tawa1 last updated on 26/Nov/18

	$Sir, does this work for every circle like this ?$
	Commented by Tawa1 last updated on 26/Nov/18

	$And please sir, reference prove of similar circle you have prove$ $here before . I want to understand the concept of the question$ $God bless you sir$
	Commented by mr W last updated on 26/Nov/18

	Commented by mr W last updated on 26/Nov/18

	$A C = r_{1} + r_{3}$ $A E = r_{1} - r_{3}$ $E C = \sqrt{{A C}^{2} - {A E}^{2}} = \sqrt{{(r_{1} + r_{3})}^{2} - {(r_{1} - r_{3})}^{2}} = \sqrt{4 r_{1} r_{3}}$ $s i m i l a r l y$ $F C = \sqrt{{B C}^{2} - {B F}^{2}} = \sqrt{{(r_{2} + r_{3})}^{2} - {(r_{2} - r_{3})}^{2}} = \sqrt{4 r_{2} r_{3}}$ $A B = r_{1} + r_{2}$ $A G = r_{1} - r_{2}$ $G B = \sqrt{{A B}^{2} - {A G}^{2}} = \sqrt{{(r_{1} + r_{2})}^{2} - {(r_{1} - r_{2})}^{2}} = \sqrt{4 r_{1} r_{2}}$ $E F = E C + F C = G B$ $\Rightarrow \sqrt{r_{1} r_{3}} + \sqrt{r_{2} r_{3}} = \sqrt{r_{1} r_{2}}$ $. . . . . .$
	Commented by Tawa1 last updated on 26/Nov/18

	$God bless you sir . I appreciate your time$
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