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Question Number 124740 by ajfour last updated on 05/Dec/20

Commented by ajfour last updated on 05/Dec/20

$F i n d e q . o f c i r c l e p a s s i n g t h r o u g h$ $i n t e r s e c t i o n p o i n t s o f c u r v e s$ $y = x^{2} - 1 a n d y = \frac{c}{x}; (c < \frac{2}{3 \sqrt{3}})$
	Answered by ajfour last updated on 05/Dec/20

	$y = x^{2} - 1$ $y = \frac{c}{x}$ $l e t r o o t s o f x^{2} - 1 = \frac{c}{x} b e p, q, r$ $p + q + r = 0$ $p q + q r + r p = - 1$ $p q r = c$ ${(y - k)}^{2} + {(x - h)}^{2} = R^{2}$ ${(x^{2} - 1 - k)}^{2} + {(x - h)}^{2} = R^{2}$ $p (q^{2} + r^{2} - 2 - k) + p + 2 h = 0$ $\Rightarrow p (p^{2} - \frac{2 c}{p} - 2 - k) + p + 2 h = 0$ $\Rightarrow 2 h - (k + 2) p + 2 p - c = 0 . . . . (I)$ $\Rightarrow Σ g i v e s$ $\Rightarrow h = \frac{c}{2}$ $A n d c o n s i d e r i n g (I) a g a i n$ $w i t h h = c \Rightarrow k = 0$ $\Rightarrow R^{2} = {(x^{2} - 1)}^{2} + {(x - \frac{c}{2})}^{2}$ $\Rightarrow R^{2} = 1 + \frac{c^{2}}{4}$ $H e n c e e q . o f c i r c l e i s$ ${(x - \frac{c}{2})}^{2} + y^{2} = 1 + \frac{c^{2}}{4}$ $e x a m p l e :$ $F o r c = \frac{1}{3}, t h e c u r v e s a r e$ $y = x^{2} - 1, y = \frac{1}{3 x}, a n d t h e c i r c l e$ ${(x - \frac{1}{6})}^{2} + y^{2} = \frac{37}{36} ★$
	Commented by ajfour last updated on 05/Dec/20

	Commented by mr W last updated on 05/Dec/20

	$f a n t a s t i c!$
	Commented by ajfour last updated on 06/Dec/20

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