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Isomorphism · #2 June 14th, 2008, 06:17 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) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$

Yes the co-ordinates are right.

But the area is wrong. The area is a constant and it is 1.

I did it the long way. First let $O_1O_2$ be the line joining the centers of $C_1$ and $C_2$ .

Then equation of the line $O_1O_2$ is $\frac{x}{t} + yt + 1 = 0$ .

Now the base of the triangle is $|O_1O_2| = \sqrt{t^2 + \frac1{t^2}}$ .

The perpendicular distance(height) from $(0,0)$ to the line is $\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}}$ . So the area of the triangle is $\frac12 |O_1O_2|\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |\frac{0}{t} + 0t + 1|= \frac12$ .

The perpendicular distance(height) from $(-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4})$ to the line is $\frac{|\frac{-\frac {2t}{1+t^4}}{t} -\frac {2t^3 }{1+t^4}t + 1|}{\sqrt{t^2 + \frac1{t^2}}}$ . So the area of the triangle is $\frac12 |O_1O_2|\frac{|-\frac {2}{1+t^4} -\frac {2t^4 }{1+t^4} + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |-\frac {2+2t^4 }{1+t^4} + 1| =\frac12 |-2 + 1| = \frac12$ .

So the area of the quadrilateral is the sum of the area of triangles.

And that is 1, a constant.

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