x^2+y^2=2ax

$${\int }_0^a\sqrt[]{a^2-x^2}dx=\frac{\pi }{4}{a}^2$$
$${\int }_0^a\sqrt[]{2ax-x^2}dx={\int }_0^a\sqrt[]{a^2-(x-a{)}^2}d(x-a)=\frac{\pi }{4}{a}^2$$