$26 \cdot 7$ Connecting 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) a line segment.
(B) a scalene triangle.
(C) an equilateral triangle.
(D) a quadrilateral.
(China High School Mathematics League, 1996)
[Solution] The coordinates of the intersection point $A(x, y)$ of the circle $x^{2}+(y-1)^{2}=1$ and the ellipse $9 x^{2}+(y+1)^{2}=9$ should satisfy the equation
$$
x^{2}+\frac{(y+1)^{2}}{9}=x^{2}+(y-1)^{2} \text {. }
$$

Solving this, we get $y=2$ or $y=\frac{1}{2}$.
Thus, the corresponding values of $x$ are $x=0$ or $x= \pm \frac{\sqrt{3}}{2}$.
Therefore, 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 $\left|A_{1} A_{2}\right|=\left|A_{1} A_{3}\right|=\left|A_{2} A_{3}\right|=\sqrt{3}$.
Thus, the resulting figure is an equilateral triangle.
Therefore, the answer is $(C)$.