$26 \cdot 5$ In $\triangle A B C$, $A B=2$ is a fixed length. From $A$, draw the median $A D$ to $B C$, $A D = 1 \frac{1}{2}$. Also, $D$ is a moving point, then the locus of vertex $C$ of the triangle is 
(A) The line $A B$, from $A$ with a length of $1 \frac{1}{2}$.
(B) A circle, with $A$ as the center, and radius 2.
(C) A circle, with $A$ as the center, and radius 3.
(D) A circle, with radius 3, and the center on $A B$, 4 units from $B$.
(E) An ellipse, with $A$ as a focus.
(1st American High School Mathematics Examination, 1950)
[Solution] Taking $A$ as the origin and the side $AB$ as the x-axis, the coordinates of $A$ and $B$ are $(0,0)$ and $(2,0)$, respectively. Let the coordinates of $C$ be $(x, y)$, then the coordinates of the midpoint $D$ of $BC$ are $\left(\frac{x+2}{2}, \frac{y}{2}\right)$. By the given condition,
$$
\begin{aligned}
A D & =\sqrt{\left(\frac{x+2}{2}-0\right)^{2}+\left(\frac{y}{2}-0\right)^{2}} \\
& =\frac{3}{2},
\end{aligned}
$$

which simplifies to $(x+2)^{2}+y^{2}=9$, representing a circle with center at $(-2,0)$ and radius 3.
Therefore, the correct choice is (D).
Note: Since $C$ is a vertex of the triangle, $C$ cannot lie on the side $AB$, so the trajectory of $C$ should be the circle in (D) excluding the points $(1,0)$ and $(-5,0)$.