10. Ans: C
$$
\begin{aligned}
x^{2}+y^{2}-2 x+2 y=2 & \Longleftrightarrow\left(x^{2}-2 x+1\right)+\left(y^{2}+2 y+1\right)=4 \\
& \Longleftrightarrow(x-1)^{2}+(y+1)^{2}=2^{2} .
\end{aligned}
$$

Therefore, the points $(x, y)$ in the Cartesian plane satisfying the equation consist of points on the circle centred at $(1,-1)$ and of radius 2 . For a point $(x, y)$ on the circle, it is easy to see that its distance $\sqrt{x^{2}+y^{2}}$ from the origin $(0,0)$ is greatest when it is the point in the fourth quadrant lying on the line passing through $(0,0)$ and $(1,-1)$. In other words,
$$
(x, y)=(1+\sqrt{2},-1-\sqrt{2}) \text {, and } x^{2}+y^{2}=(1+\sqrt{2})^{2}+(-1-\sqrt{2})^{2}=6+4 \sqrt{2} \text {. }
$$
10. Let $x$ and $y$ be real numbers such that
$$
x^{2}+y^{2}=2 x-2 y+2 .
$$

What is the largest possible value of $x^{2}+y^{2}$ ?
(A) $10+8 \sqrt{2}$
(B) $8+6 \sqrt{2}$
(C) $6+4 \sqrt{2}$
(D) $4+2 \sqrt{2}$
(E) None of the above