Hello arze
\(\displaystyle A(x_1, y_1)\) and \(\displaystyle B(x_2, y_2)\) are the ends of the diameter of a circle. Show that the equation of the circle is:

\(\displaystyle (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=1\).

If A and B are also points on the hyperbola \(\displaystyle xy=c^2\) and the circle on AB as diameter cuts the hyperbola again at C and D, prove that CD is a diameter of the hyperbola.

I can't prove the first part. I got:

\(\displaystyle (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\), instead of the given equation. I found this from the midpoint of AB and half the length of AB.

Thanks!

You are right: the equation of the circle is \(\displaystyle (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\). You can prove it in two lines.

Let \(\displaystyle P\;(x,y)\) be any point on the circle with diameter AB. Then, since the angle in a semicircle is a right-angle, AP and BP are perpendicular. So:\(\displaystyle \frac{y-y_1}{x-x_1}\cdot \frac{y-y_2}{x-x_2} = -1\)

\(\displaystyle \Rightarrow(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\)

For the second part, note first that the rectangular hyperbola \(\displaystyle xy=c^2\) has rotational symmetry about the origin. Therefore the coordinates of the ends of a diameter will be of the form \(\displaystyle (\pm h,\pm k)\). So, in order to prove that the other points of intersection are at opposite ends of a diameter, all we need show is that their \(\displaystyle x\)-coordinates (and therefore their \(\displaystyle y\)-coordinates) differ only in sign.

Now for any point \(\displaystyle (h,k)\) on the hyperbola:\(\displaystyle k = \frac{c^2}{h}\) ...(1)

and, in particular:\(\displaystyle y_1 = \frac{c^2}{x_1}\) ...(2)

and\(\displaystyle y_2 = \frac{c^2}{x_2}\) ...(3)

If \(\displaystyle (h,k)\) also lies on the circle, then:\(\displaystyle (h-x_1)(h-x_2)+(k-y_1)(k-y_2)=0\)

So, from (1), (2) and (3), \(\displaystyle h\) satisfies the equation:\(\displaystyle (h-x_1)(h-x_2)+\left(\frac{c^2}{h}-\frac{c^2}{x_1}\right)\left(\frac{c^2}{h}-\frac{c^2}{x_2}\right)=0\)

\(\displaystyle \Rightarrow (h-x_1)(h-x_2)+\frac{c^4}{x_1x_2h^2}(x_1-h)(x_2-h)=0\)

\(\displaystyle \Rightarrow (h-x_1)(h-x_2)\left(1+\frac{c^4}{x_1x_2h^2}\right)=0\)

So the four roots of this equation are given by:\(\displaystyle h = x_1, \;x_2,\; \pm\frac{c^2}{\sqrt{-x_1x_2}}\)

The last two roots differ only in sign, thus proving that the third and fourth points of intersection lie at the ends of a diameter of the hyperbola.

Grandad