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Question Number 15480 by ajfour last updated on 10/Jun/17

Commented by ajfour last updated on 10/Jun/17

$Q . 15434 a n s w e r e d h e r e .$ $T o f i n d m a x i m u m l e n g t h o f$ $A B, i n t e r m s o f r, R, a n d d .$
	Answered by ajfour last updated on 11/Jun/17

	$T o f i n d P (h, k) .$ $x^{2} + y^{2} = r^{2}$ ${(x - d)}^{2} + y^{2} = R^{2}$ $s u b t r a c t i n g t o o b t a i n h$ $2 d h = R^{2} - r^{2}$ $h = \frac{R^{2} - r^{2}}{2 d}$ $k = \sqrt{r^{2} - h^{2}} (c o n s i d e r i n g j u s t$ $t h e p o s i t i v e o n e h e r e)$ $l e t e q u a t i o n o f A P B b e,$ $y = m (x - h) + k$ $s o l v i n g t h i s w i t h x^{2} + y^{2} = r^{2}$ $x^{2} + {[m (x - h) + k]}^{2} - r^{2} = 0$ $(1 + m^{2}) x^{2} + (- 2 m^{2} h + 2 m k) x + c_{1} = 0$ $i t s r o o t s a r e h a n d x_{1} .$ $s o x_{1} + h = - (\frac{- 2 m^{2} h + 2 m k}{1 + m^{2}}) . . . (i)$ $A g a i n s o l v i n g l i n e$ $y = m (x - h) + k$ $w i t b {(x - d)}^{2} + y^{2} = R^{2}$ ${(x - d)}^{2} + {[m (x - h) + k]}^{2} - R^{2} = 0$ $(1 + m^{2}) x^{2} + (- 2 d - 2 m^{2} h + 2 m k) x + c_{2} = 0$ $i t s r o o t s a r e h, a n d x_{2} .$ $x_{2} + h = - (\frac{- 2 d - 2 m^{2} h + 2 m k}{1 + m^{2}}) . . (i i)$ $e q n . (i) - e q n . (i i) g i v e s :$ $x_{2} - x_{1} = \frac{2 d}{1 + m^{2}}$ $l e n g t h o f A B = \sqrt{1 + m^{2}} (x_{2} - x_{1})$ $l_{A B} = \frac{2 d}{\sqrt{1 + m^{2}}} a n d i s c l e a r l y$ $m a x i m u m f o r m = 0, a n d$ ${(l e n g t h A B)}_{m a x} = 2 d .$
	Commented by mrW1 last updated on 11/Jun/17

	$V ERY NICE$
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