Let A and B be two points on the hyperbola xy=1, and let C be the reflection of B through the origin.

Let Gamma be the circumcircle of triangle ABC and let A' be the point on Gamma diametrically opposite A. Show that A' is also on the hyperbola xy=1.

Nvrm I solved it.

To show that point A' is also on the hyperbola xy=1, we need to prove that the coordinates of A' satisfy the equation.

Let's start by finding the coordinates of A, B, and C.

From the equation of the hyperbola xy=1, we can solve for y in terms of x: y = 1/x.

Point A lies on the hyperbola, so let's assume its x-coordinate is a. Therefore, its y-coordinate is 1/a. So, the coordinates of A are (a, 1/a).

The reflection of a point (x, y) through the origin is (-x, -y). So, the coordinates of C, the reflection of B through the origin, can be found by taking the negative of the coordinates of B.

Let's assume the x-coordinate of B is b. Therefore, its y-coordinate is 1/b. So, the coordinates of B are (b, 1/b).

The coordinates of C, the reflection of B through the origin, are (-b, -1/b).

Next, we need to find the equation of the circumcircle of triangle ABC.

The circumcircle of a triangle is the circle that passes through all three vertices of the triangle.

The equation of a circle with center (h, k) and radius r is given by: (x - h)^2 + (y - k)^2 = r^2.

Let's suppose the center of the circumcircle of triangle ABC is (h, k) and the radius is r.

We know that A, B, and C lie on the circumcircle, so they satisfy the equation of the circle:

(a - h)^2 + (1/a - k)^2 = r^2, (b - h)^2 + (1/b - k)^2 = r^2, (-b - h)^2 + (-1/b - k)^2 = r^2.

Simplifying these equations, we get:

a^2 - 2ah + h^2 + (1/a)^2 - 2(1/a)k + k^2 = r^2, b^2 - 2bh + h^2 + (1/b)^2 - 2(1/b)k + k^2 = r^2, b^2 + 2bh + h^2 + (1/b)^2 + 2(1/b)k + k^2 = r^2.

Canceling out the common terms, we obtain:

a^2 - 2ah + (1/a)^2 - 2(1/a)k = 0, b^2 - 2bh + (1/b)^2 - 2(1/b)k = 0, b^2 + 2bh + (1/b)^2 + 2(1/b)k = 0.

Now, let's find the coordinates of A', the point diametrically opposite A on the circumcircle.

Since A' is diametrically opposite A, the x-coordinate of A' is the negative of the x-coordinate of A, and the y-coordinate of A' is the negative of the y-coordinate of A.

Therefore, the coordinates of A' are (-a, -1/a).

To show that A' lies on the hyperbola xy=1, we substitute the coordinates of A' into the equation:

(-a)(-1/a) = 1.

This simplifies to:

1 = 1,

which is true.

Therefore, we have shown that the point A' is also on the hyperbola xy=1.

the hyperbola xy=1 has two branches, symmetric about the origin, since (-x)(-y) = 1

So, if A = (x,y), then A'=(-x,-y) and is also on the hyperbola.
There is no need to introduce any circle, since A' is directly opposite A.