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Question Number 158961 by ajfour last updated on 11/Nov/21

Commented by ajfour last updated on 11/Nov/21

$F i n d r g i v e n, c .$
Commented by 1549442205PVT last updated on 11/Nov/21

$y o u r c o n d i t o n {i s n}^{'} t s u i t a b l e . S o i f t h a t$ $r^{2} + x^{2} = r^{2} - 1 + 1 = r^{2} \Rightarrow x = 0 ?$
Commented by 1549442205PVT last updated on 13/Nov/21

$E c x c u s e m e, i m i s t a k e! T h a n k y o u s i r$ $f o r i n t e r e s t i n g p r o b l e m s .$
	Answered by ajfour last updated on 12/Nov/21

	$(r^{2} - 1) \sqrt{r^{2} - 1} + \sqrt{r^{2} - 1}$ $= r^{2} \sqrt{{(r + c)}^{2} - r^{4}}$ $\Rightarrow r^{4} + r^{2} = 1 + {(r + c)}^{2}$ $\Rightarrow \frac{r^{6} - 1}{r^{2} - 1} = 2 + {(r + c)}^{2}$ $r^{6} - 2 r^{2} + 1 = (r^{2} - 1) {(r + c)}^{2}$ ${(r + c)}^{2} - 2 (r^{2} - 1)$ $= 1 + (r^{2} - 1) {(r + c)}^{2}$ $\Rightarrow 2 + \frac{1}{{(r + c)}^{2}} = \frac{2 r^{2}}{{(r + c)}^{2}} + r^{2}$ $. . . . . .$
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