1. The common points of the circle $x^{2}+(y-1)^{2}=1$ and the ellipse $9 x^{2}+(y+1)^{2}=9$, when connected by line segments, form the shape of
(A) a line segment;
(B) a scalene triangle;
(C) an equilateral triangle;
(D) a quadrilateral.
1. (C)

The circle $x^{2}+(y-1)^{2}=1$ and the ellipse $9 x^{2}+(y+1)^{2}=9$ intersect at points $A(x, y)$ whose coordinates satisfy the equation, $x^{2}+\frac{(y+1)^{2}}{9}=x^{2}+(y-1)^{2}$.

Solving this, we get $y=2$, or $y=\frac{1}{2}$. Thus, the corresponding values of $x$ are $x=0$, or $x= \pm \frac{\sqrt{3}}{2}$. This gives us three intersection points of the given circle and ellipse: $A_{1}(0,2), A_{2}\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, $A_{3}\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. It is easy to find that, $\left|A_{1} A_{2}\right|=\left|A_{1} A_{3}\right|=\left|A_{2} A_{3}\right|=\sqrt{3}$.
Therefore, the resulting figure is an equilateral triangle, hence the answer is (C).