If a circle and the rectangular hyperbola $x y = c^{2}$ meet in the four points $p_{1} = 1, p_{2} = 2, p_{3} = 3$ and $p_{4}$ , and the center of the circle through the points $p_{1}, p_{2}$ and $p_{3}$ is $(\frac{37}{12}, \frac{47}{12})$ (where $p$ is the parameter of the parametric coordinates of the hyperbola), find $c$ .

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Solution

Let the rectangular hyperbola be $x y = c^{2}$ and the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + k = 0.$
Any point on the hyperbola is $(c p, \frac{c}{p})$ . If it lies on the circle, then $c^{2} p^{2} + \frac{c^{2}}{p^{2}} + 2 g c p + 2 f \frac{c}{p} + k = 0.$
$c^{2} p^{4} + 2 g c p^{3} + k p^{2} + 2 f c p + c^{2} = 0.$
This is fourth degree equation in $p$ , which has four roots. Hence the circle and the hyperbola intersect in four points.
If $p_{1}, p_{2}, p_{3}, p_{4}$ are the roots of this equation, then
$p_{1} + p_{2} + p_{3} + p_{4} = \frac{2 g c}{c^{2}} = \frac{2 g}{c}$
$c p_{1} + c p_{2} + c p_{3} + c p_{4} = - 2 g$
$\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = g / 2$ ..(i)
Also $\frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{p_{3}} + \frac{1}{p_{4}} = \frac{p_{1} . p_{2} . p_{3}}{p_{1} . p_{2} . p_{3} . p_{4}} = \frac{2 f c}{c^{2}} = \frac{2 f}{c}$
$\frac{c}{p_{1}} + \frac{c}{p_{2}} + \frac{c}{p_{3}} + \frac{c}{p_{4}} = 2 f$
$\frac{y_{1} + y_{2} + y_{3} + y_{4}}{4} = \frac{f}{2}$ ..(ii)
Also, $p_{1} . p_{2} . p_{3} . p_{4} = \frac{c^{2}}{c^{2}} = 1$ ..(iii)
Hence from (i),(ii) and (iii) we can clearly see that
$- g, - f = \frac{c}{2} (p_{1} + p_{2} + p_{3} + \frac{1}{p_{1} . p_{2} . p_{3}}), \frac{c}{2} (\frac{1}{p_{1}} + \frac{1}{p_{2}} + \frac{1}{p_{3}} + p_{1} . p_{2} . p_{3})$
Hence, comparing the centers, we get $c = 1.$

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If a circle and the rectangular hyperbola xy=c2 meet in the four points p1=1,p2=2,p3=3 and p4, and the center of the circle through the points p1,p2 and p3 is (3712,4712) (where p is the parameter of the parametric coordinates of the hyperbola), find c.