Equation of circle and diameter of hyperbola.

Jul 2009
338
14
Singapore
\(\displaystyle A(x_1, y_1)\) 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
May 2006
12,028
6,341
Lexington, MA (USA)
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" . . .


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


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


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


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

 
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Jul 2009
338
14
Singapore
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!
 
Jun 2009
806
275
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.
 
Jul 2009
338
14
Singapore
AB is a diameter of the curve, so the points are in the first and third quadrants
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
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
 
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