a(x^2 +y^2 )+b(x+y)= c & x^2 −y^2 = R^2 Solve for x or y .


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Question Number 47966 by ajfour last updated on 17/Nov/18

$a (x^{2} + y^{2}) + b (x + y) = c$ $& x^{2} - y^{2} = R^{2}$ $S o l v e f o r x o r y .$
Commented bybehi83417@gmail.com last updated on 17/Nov/18

$x + y = t, x - y = s \Rightarrow 4 x y = t^{2} - s^{2}$ $a (t^{2} - \frac{t^{2} - s^{2}}{2}) + b t = c, t s = R^{2}$ $\Rightarrow a (t^{2} + s^{2}) + 2 b t = 2 c \Rightarrow a (t^{2} + \frac{R^{4}}{t^{2}}) + 2 b t = 2 c$ $\Rightarrow {a t}^{4} + 2 {b t}^{3} + 2 {c t}^{2} + {a R}^{4} = 0$ $\Rightarrow t^{4} + \frac{2 b}{a} t^{3} + \frac{2 c}{a} t^{2} + R^{4} = 0$ $b y h a v i n g n u m e r i c v a l v e s, w e c a n$ $s o l v e t h i s f o r t a n d t h e n s .$
	Answered by MJS last updated on 17/Nov/18

	$x + y = \frac{c - a (x^{2} + y^{2})}{b}$ $x + y = \frac{R^{2}}{x - y}$ $\frac{c - a (x^{2} + y^{2})}{b} = \frac{R^{2}}{x - y}$ $x^{3} - {y x}^{2} + \frac{{a y}^{2} - c}{a} x + \frac{{b R}^{2} - y ({a y}^{2} - c)}{a} = 0$ $x = z + \frac{y}{3}$ $z^{3} + \frac{2 {a y}^{2} - 3 c}{3 a} z + \frac{27 {b R}^{2} - 2 y (10 {a y}^{2} - 9 c)}{27 a} = 0$ $and this can be solved . . .$
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