1. Connect the common points of the circle $x^{2}+(y-1)^{2}=1$ and the ellipse $9 x^{2}+(y+1)^{2}$ $=9$ with line segments, the resulting figure is
(A) line segment
(B) scalene triangle
(C) equilateral triangle
(D) quadrilateral
-、(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 must satisfy the equation
$$
x^{2}+\frac{(y+1)^{2}}{9}=x^{2}+(y-1)^{2} \text {. }
$$

Solving this, we get $y=2$ or $\frac{1}{2}$.
Thus, the corresponding values of $x$ are $0$ or $x= \pm \frac{\sqrt{3}}{2}$.
Hence, we obtain 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.