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Question Number 53252 by ajfour last updated on 19/Jan/19

Commented by ajfour last updated on 19/Jan/19

$F i n d a a n d r i n t e r m s o f b .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

	$c e n t r e A i s {(x - 0)}^{2} + {(y - b)}^{2} = a^{2}$ $c e n t r e B i s {(x - a - b)}^{2} + {(y - b)}^{2} = b^{2}$ $A C = \sqrt{{(r - 0)}^{2} + {(r - b)}^{2}} = r + a$ $r^{2} + r^{2} - 2 r b + b^{2} = r^{2} + 2 r a + a^{2}$ $r^{2} - 2 r b + b^{2} - 2 r a - a^{2} = 0 . . . . e q n 1$ $B C$ ${(a + b - r)}^{2} + {(b - r)}^{2} = {(r + b)}^{2}$ $a^{2} + b^{2} + r^{2} + 2 a b - 2 a r - 2 b r + b^{2} - 2 b r + r^{2} = r^{2} + 2 b r + b^{2}$ $a^{2} + 2 b^{2} + 2 r^{2} + 2 a b - 2 a r - 4 b r - r^{2} - 2 b r - b^{2} = 0$ $a^{2} + b^{2} + r^{2} + 2 a b - 2 a r - 6 b r = 0 . . . . e q n 2$ $r^{2} - 2 r a - a^{2} = 2 r b - b^{2} f r l m 1$ $r^{2} - 2 a r + a^{2} = 6 b r - 2 a b - b^{2} f r o m 2$ $r^{2} - 2 r a + b^{2} = a^{2} + 2 r b$ $r^{2} - 2 r a + b^{2} = 6 b r - 2 a b - a^{2}$ $s o a^{2} + 2 r b = 6 b r - 2 a b - a^{2}$ $2 a^{2} - 4 b r + 2 a b = 0$ $r = \frac{2 a^{2} + 2 a b}{4 b} = \frac{a (a + b)}{2 b} \leftarrow$ $r^{2} - 2 a r - a^{2} = 2 r b - b^{2}$ ${(\frac{a^{2} + a b}{2 b})}^{2} - 2 (a + b) (\frac{a^{2} + a b}{2 b}) = a^{2} - b^{2}$ $\frac{a^{2} {(a + b)}^{2}}{4 b^{2}} - \frac{(a + b) (a + b) a}{b} = (a + b) (a - b)$ $d e v i d i n g b y a + b b o t h s i d e$ $\frac{a^{2} (a + b)}{4 b^{2}} - \frac{a (a + b)}{b} = a - b$ $(a + b) (\frac{a^{2}}{4 b^{2}} - \frac{a}{b}) = a - b$ $\frac{a^{2}}{4 b^{2}} - \frac{a}{b} = \frac{a - b}{a + b}$ $l e t \frac{a}{b} = k$ $\frac{k^{2}}{4} - k = \frac{k - 1}{k + 1}$ $\frac{k^{2} - 4 k}{4} = \frac{k - 1}{k + 1}$ $k^{3} - 4 k^{2} + k^{2} - 4 k = 4 k - 4$ $k^{3} - 3 k^{2} - 8 k + 4 = 0$
	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

	$k = \frac{a}{b} = 0.438 s o a = 0.438 b$ $\frac{a}{b} = 4.562 a = 4.562 b \to t h i s s o l u t i o n n o t f e a s i b l e$ $f r o m d i a g r a m . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

	$r = \frac{a (a + b)}{2 b} = \frac{0.438 b (0.438 b + b)}{2 b} = 0.219 (1.438 b)$ $r = 0.3149 b$
	Answered by ajfour last updated on 19/Jan/19

	$i o b t a i n$ $a = b (\sqrt{2} - 1)$ $r = (2 + \sqrt{2} - 2 \sqrt{1 + \sqrt{2}}) b .$
	Answered by ajfour last updated on 19/Jan/19

	$\underline{F r o m △ A E C}$ ${(b - r)}^{2} + r^{2} = {(a + r)}^{2} . . . (i)$ $B C^{2} = {(b + r)}^{2} = {(a + b - r)}^{2} + {(b - r)}^{2}$ $. . . . (i i)$ $\Rightarrow r^{2} - 2 (a + b) r + b^{2} - a^{2} = 0 &$ $r^{2} - 2 (a + 3 b) r + {(a + b)}^{2} = 0$ $c o m p a r i n g t h e t w o$ $\frac{a + b}{a + 3 b} = \frac{b - a}{a + b}$ $\Rightarrow a^{2} + 2 a b + b^{2} = 3 b^{2} - a^{2} - 2 a b$ $o r a^{2} + 2 a b + b^{2} = 2 b^{2}$ $\Rightarrow a + b = b \sqrt{2}$ $a = (\sqrt{2} - 1) b .$ $n o w p i c k i n g u p t h e 2^{n d} q u a d r a t i c i n r$ $r^{2} - 2 (a + 3 b) r + {(a + b)}^{2} = 0$ $b u t w e f o u n d a = \sqrt{2} b - b, s o$ $r^{2} - 2 (b \sqrt{2} - b + 3 b) r + 2 b^{2} = 0$ $r^{2} - 2 (2 + \sqrt{2}) b + 2 b^{2} = 0$ $\Rightarrow r = (2 + \sqrt{2} \pm \sqrt{{(2 + \sqrt{2})}^{2} - 2}) b$ $f e a s i b l e o n e f o r d i a g r a m i s$ $r = (2 + \sqrt{2} - 2 \sqrt{1 + \sqrt{2}}) b .$
	Commented by mr W last updated on 20/Jan/19

	$s i r, i {d o n}^{'} t t h i n k w e c a n g e t$ $\frac{a + b}{a + 3 b} = \frac{b - a}{a + b}$ $f r o m$ $r^{2} - 2 (a + b) r + b^{2} - a^{2} = 0 &$ $r^{2} - 2 (a + 3 b) r + {(a + b)}^{2} = 0 .$ ${l e t}^{'} s l o o k a t$ $a_{1} x^{2} + b_{1} x + c_{1} = 0$ $a_{2} x^{2} + b_{2} x + c_{2} = 0$ $i f b o t h e q n . h a v e t w o e q u a l r o o t s,$ $t h e n$ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ $i f b o t h e q n . h a v e o n l y o n e e q u a l r o o t,$ $w e c a n n o t s a y$ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$ $i t h i n k w e h a v e h e r e t h e s e c o n d c a s e .$
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