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Question Number 213818 by ajfour last updated on 17/Nov/24

Commented by ajfour last updated on 17/Nov/24

$F i n d R i n t e r m s s m a l l e r r a d i i a, b .$
	Answered by mr W last updated on 17/Nov/24

	Commented by ajfour last updated on 17/Nov/24

	$x^{2} + y^{2} = R^{2}$ ${(x - R - r)}^{2} + y^{2} = r^{2}$ $s l o p e o f c o m m o n t a n g e n t$ $m^{2} = \frac{{(R - a)}^{2}}{4 R a} = \frac{4 R b}{{(R - b)}^{2}}$ $R^{2} - (a + b) R + a b = 4 R \sqrt{a b}$ $R = \frac{a + b}{2} + 2 \sqrt{a b} + \sqrt{{(\frac{a + b}{2} + 2 \sqrt{a b})}^{2} - a b}$ $s a y b = 1, a = 2$ $R = \frac{3}{2} + 2 \sqrt{2} + \sqrt{\frac{9}{4} + 8 + 6 \sqrt{2} - 2}$ $R = \frac{3}{2} + 2 \sqrt{2} + \sqrt{\frac{33}{4} + 6 \sqrt{2}}$ $\approx 8.4193$
	Commented by mr W last updated on 17/Nov/24

	$y e s . n o w w e g o t t h e s a m e .$
	Commented by mr W last updated on 18/Nov/24

	$\sqrt{{(R + a)}^{2} - {(R - a)}^{2}} + \sqrt{{(R + b)}^{2} - {(R - b)}^{2}} = \sqrt{{(R + a)}^{2} + {(R + b)}^{2} - {(a - b)}^{2}}$ $\sqrt{2 a R} + \sqrt{2 b R} = \sqrt{(R + a) (R + b)}$ $R^{2} - (a + b + 4 \sqrt{a b}) R + a b = 0$ $R = \frac{a + b}{2} + 2 \sqrt{a b} + \sqrt{(\frac{a + b}{2} + \sqrt{a b}) (\frac{a + b}{2} + 3 \sqrt{a b})}$
	Commented by ajfour last updated on 17/Nov/24

	$p l z e x p l a i n s i r, R . H . S . o f f i r s t l i n e . .$
	Commented by mr W last updated on 17/Nov/24

	${A B}^{2} = {(\sqrt{{(R + a)}^{2} - {(R - a)}^{2}} + \sqrt{{(R + b)}^{2} - {(R - b)}^{2}})}^{2} + {(a - b)}^{2}$ ${A B}^{2} = {(R + a)}^{2} + {(R + b)}^{2}$
	Commented by mr W last updated on 17/Nov/24

	Commented by ajfour last updated on 17/Nov/24

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