I'm aware that one can prove the area of a circle is $πr^2$ by integration, summing vertical strips of a quarter of a circle. 
Starting at the origin, the height of each "strip" is $y = \sqrt{r^2 - x^2}$ and the width is $dx$ giving $$ A = 4\int_0^r \sqrt{r^2 - x^2} \,\,dx$$ After integrating I believe one eventually gets $$4[r^2arctan(1) + r · 0 − r^2arctan(0) − 0] = 4(\frac{1}{4}πr^2) $$
My question is this: Suppose you didn't know the formula for a circle was $y^2 + x^2 = r^2$ but reasoned as follows:
Instead of working with the piece of the circle in quadrant I, we decide to sum over the piece in quadrant II, but shift the circle to the right for ease of computation.
Now, we once again integrate over $0$ to $r$ but since we don't know the equation of a circle in order to find height $y$ we reason by the Pythagorean theorem that $y = \sqrt{r^2 - (r- x)^2} = \sqrt{-2rx +x^2}$
Therefore the area of the circle should be $$A = 4\int_0^r \sqrt{-2rx +x^2} \,\,dx$$
But based on the antiderivative of the expression alone (according to software), this doesn't appear to be right.
I'm prone to careless oversights, but otherwise please let me know where this argument goes wrong.