4. 1. 10 *When $a$ takes what value, the system of inequalities $\left\{\begin{array}{l}x^{2}+y^{2}+2 x \leqslant 1, \\ x-y+a=0\end{array}\right.$, has a unique solution, and find it.
From (2), we get $y=x+a$. Substituting into (1) yields $2 x^{2}+2(a+1) x+a^{2}-1 \leqslant 0$. Let $f(x)=2 x^{2}+2(a+1) x+a^{2}-1$, then $f(x)$ is a parabola opening upwards. For $f(x) \leqslant 0$ to have a unique solution, the parabola must intersect the $x$-axis at exactly one point, hence $\Delta=4(a+1)^{2}-8\left(a^{2}-1\right)=0$, so $a^{2}-2 a-3=0$. Solving this gives $a=3, a=-1$.
When $a=3$, the unique solution to (3) is $x=-2$. Substituting into (2) gives $y=1$.
When $a=-1$, the unique solution to (3) is $x=0$. Substituting into (2) gives $y=-1$.
In summary, when $a=3$, the unique solution is $(x, y)=(-2,1)$; when $a=-1$, the unique solution is $(x, y)=(0,-1)$.