It isn't necessary to calculate the equation of the circle, nor even the co-ordinates of its centre.
Melody calculated the co-ordinates of the centre as (a, (9a - 16)/3), but to save on the typing I'll use (a, A).
All we need do is to find two other points on the hyperbola that are the same distance from the centre as the two given points.
Let the co-ordinates of a general point on the hyperbola be (p, 1/p), then we need, (using the point (1, 1) as one of the given points),
(p - a)^2 + (1/p - A)^2 = (1 - a)^2 + (1 - A)^2,
multiplying that out,
p^2 + a^2 -2ap + (1/p)^2 + A^2 - 2A/p = 1 + a^2 -2a + 1 + A^2 - 2A,
cancelling the a^2 and the A^2 and then multiplying throughout by p^2 gets you
p^4 - 2ap^3 + 2(a + A - 1)p^2 - 2Ap + 1 = 0.
Suppose that the roots of this are r, s, t, u, so that
(p - r)(p - s)(p - t)(p - u) = 0,
then, after expanding (which you don't need to do !) and equating constants with the actual equation, rstu = 1.
However, two of the roots are known, they're 1 and 3, so the product of the other two must be 1/3.
That means that the product of the other two y co-ordinates will be 3.
Tiggsy
(Sorry, I couldn't get LaTeX to work for some reason.)
