# Equation of circle and diameter of hyperbola.

#### 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!

#### Soroban

MHF Hall of Honor
Hello, arze!

There is a typo in the problem . . .

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

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

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

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

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

And the rest is "just Algebra" . . .

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

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

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

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

• arze

#### arze

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

#### sa-ri-ga-ma

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.

#### arze

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

MHF Hall of Honor
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.

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