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Dodgy step in the Far field approximation

The Fresnel diffraction integral is:

$A(x_0 , y_0 ) = \frac{i e^{-ikz}}{λz} \int \int dx dy A( x , y ) e^{\frac{-ik}{2z} [(x - x_0)^2 + (y - y_0)^2]}$

When we want to obtain the Fraunhofer diffraction integral from here, we need to somehow convert it to:

$A(x_0 , y_0 ) = \frac{i e^{-ikz}}{λz} \int \int dx dy A( x , y ) e^{\frac{+ik}{z} [x x_0 + y y_0]}$

So I thought we should do it as follows:

$\frac{-ik}{2z} [(x - x_0)^2 + (y - y_0)^2] = \frac{-ik}{2z} [x^2 + x_0^2 + y^2 + y_0^2 - 2x x_0 - 2y y_0 ]$

And then it seems that we should neglect: $x^2 + x_0^2 + y^2 + y_0^2$ since they're all much smaller than z.
Then we get the correct solution.

But I don't see why we could do that, and leave out the $- 2x x_0 - 2y y_0$. After all they are of the same order... Please help!
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