8. The polar equation of a circle is $\rho=\sqrt{2}(\cos \theta+\sin \theta)$, the coordinates of the center of the circle are ( ).
A. $\left(2, \frac{\pi}{4}\right)$
B. $\left(1, \frac{\pi}{4}\right)$
C. $\left(1, \frac{3 \pi}{4}\right)$
D. $\left(\sqrt{2},-\frac{\pi}{4}\right)$
8. Transform the equation of the circle into Cartesian coordinates, we get $x^{2}+y^{2}-\sqrt{2} x-\sqrt{2} y=0$, which is $\left(x-\frac{\sqrt{2}}{2}\right)^{2}+\left(y-\frac{\sqrt{2}}{2}\right)^{2}=1$, with the center at $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ and radius 1, converting to polar coordinates gives $\left(1, \frac{\pi}{4}\right)$. Therefore, the answer is B. Or from $\rho=$ $\sqrt{2}(\cos \theta+\sin \theta)=2 \cos \left(\theta-\frac{\pi}{4}\right)$, we get the center at $\left(1, \frac{\pi}{4}\right)$.