# Thread: Equation of circle and diameter of hyperbola.

1. ## Equation of circle and diameter of hyperbola.

$A(x_1, y_1)$ and $B(x_2, y_2)$ are the ends of the diameter of a circle. Show that the equation of the circle is:
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=1$.
If A and B are also points on the hyperbola $xy=c^2$ 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:
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$, instead of the given equation. I found this from the midpoint of AB and half the length of AB.
Thanks!

2. Hello, arze!

There is a typo in the problem . . .

$A(h_1, k_1)$ and $B(h_2, k_2)$ are the ends of the diameter of a circle.

Show that the equation of the circle is: . $(x-h_1)(x-h_2)+(y-k_1)(y-k_2)\;=\;{\color{red}0}$
The center of the circle is: . $\left(\frac{h_1+h_2}{2},\:\frac{k_1+k_2}{2}\right)$

The diameter is: . $AB \:=\:\sqrt{(h_2-h_1)^2 + (k_2-k_1)^2}$

The radius is: . $r \;=\;\frac{1}{2}\sqrt{(h_2-h_1)^2 + (k_2-k_1)^2}$

The equation of the circle is: . $\left(x - \frac{h_1+h_2}{2}\right)^2 + \left(y - \frac{k_1+k_2}{2}\right)^2 \;=\;\frac{(h_2-h_1)^2 + (k_2-k_1)^2}{4}$

And the rest is "just Algebra" . . .

. - - - . . . $x^2 - (h_1+h_2)x + \frac{(h_1+h_2)^2}{4} + y^2 - (k_1+k_2)y + \frac{(k_1+k_2)^2}{4} \;\;=$ . $\frac{(h_2-h_1)^2}{4} + \frac{(k_2-k_1)^2}{4}$

$x^2 - (h_1+h_2)x + \frac{h_1^2}{4} + \frac{h_1h_2}{2} + \frac{h_2^2}{4} + y^2 - (k_1+k_2)y + \frac{k_1^2}{4} + \frac{k_1k_2}{2} + \frac{k_2^2}{4} \;\;=$ . $\frac{h_2^2}{4} - \frac{h_1h_2}{2} + \frac{h_1^2}{4} + \frac{k_2^2}{4} - \frac{k_1k_2}{2} + \frac{k_1^2}{4}$

. . . . . . . . . . . $\bigg[x^2 - (h_1+h_2)x + h_1h_2\bigg] + \bigg[y^2 - (k_1+k_2)y + k_1k_2\bigg] \;\;=\;\;0$

. . . . . . . . . . . . . . . . . . . . . $(x-h_1)(x-h_2) + (y-k_1)(y-k_2) \;\;=\;\;0$

3. any hints on the second part? I tried substituting $xy=c^2$ into the equation but i can't solve for x after simplifying.
Thanks!

4. If A and B are also points on the hyperbola 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 am not able to visualize the above part. The hyperbola xy = c^2 lie in the first and third quadrant. If A and B lie in the first quadrant, the circle on AB as diameter cannot cut the same hyperbola again.

5. AB is a diameter of the curve, so the points are in the first and third quadrants

6. Hello arze
Originally Posted by arze
$A(x_1, y_1)$ and $B(x_2, y_2)$ are the ends of the diameter of a circle. Show that the equation of the circle is:
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=1$.
If A and B are also points on the hyperbola $xy=c^2$ 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:
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$, 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 $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. You can prove it in two lines.

Let $P\;(x,y)$ 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:
$\frac{y-y_1}{x-x_1}\cdot \frac{y-y_2}{x-x_2} = -1$
$\Rightarrow(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
For the second part, note first that the rectangular hyperbola $xy=c^2$ has rotational symmetry about the origin. Therefore the coordinates of the ends of a diameter will be of the form $(\pm h,\pm k)$. 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 $x$-coordinates (and therefore their $y$-coordinates) differ only in sign.

Now for any point $(h,k)$ on the hyperbola:
$k = \frac{c^2}{h}$ ...(1)
and, in particular:
$y_1 = \frac{c^2}{x_1}$ ...(2)
and
$y_2 = \frac{c^2}{x_2}$ ...(3)
If $(h,k)$ also lies on the circle, then:
$(h-x_1)(h-x_2)+(k-y_1)(k-y_2)=0$
So, from (1), (2) and (3), $h$ satisfies the equation:
$(h-x_1)(h-x_2)+\left(\frac{c^2}{h}-\frac{c^2}{x_1}\right)\left(\frac{c^2}{h}-\frac{c^2}{x_2}\right)=0$

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

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

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