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Math Help - Equation of circle and diameter of hyperbola.

  1. #1
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    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!
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  2. #2
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    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

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  3. #3
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    any hints on the second part? I tried substituting xy=c^2 into the equation but i can't solve for x after simplifying.
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  4. #4
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    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.
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  5. #5
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    AB is a diameter of the curve, so the points are in the first and third quadrants
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  6. #6
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    Hello arze
    Quote Originally Posted by arze View Post
    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.

    Grandad
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