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earboth · #5 June 14th, 2008, 07:15 AM

Quote:

Originally Posted by afeasfaerw23231233

bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) ...
b) find that the area of the quadrilateral is a constant.
...

Here is a completely different attempt:

1. Calculate the coordinates of the centers of the 2 circles:

$x^2+2tx+{\color{red}t^2}+y^2 = {\color{red}t^2} ~\iff~ (x+t)^2+y^2 = t^2$

$x^2+y^2 +\frac {2y}t +{\color{red} \left(\frac1t \right)^2}={\color{red} \left(\frac1t \right)^2}~\iff~ x^2+\left(y+\frac1t\right)^2= \left(\frac1t \right)^2$

Then the area of the quadrilateral consists of 2 congruent right triangles. The legs of one of these right triangles have the lengthes(?) $R = t$ and $r = \frac1t$ .

Therefore the area of the quadrilateral is:

$A = 2 \cdot \frac12 \cdot t \cdot \frac1t = 1$

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