27 . The center of the circle passing through the points $(4,5),(6,3)$ and $(2,1)$ is ).
(A) $(8,7)$
(B) $(4.3)$
(C) $\left(\frac{13}{5}, \frac{8}{5}\right)$
(D) $\left(\frac{11}{3}, \frac{8}{3}\right)$
(E) *
27. (D).

Let $A(4,5), B(6,3), C(2,1)$. It is easy to see that the midpoint of $B C$ is $(4,2)$, and the slope is $\frac{1}{2}$. Therefore, the perpendicular bisector of $B C$ is
$$
y-2=-2(x-4) .
$$

We also know that the midpoint $D$ of $A B$ is $(5,4)$, and $A C=B C$, so the perpendicular bisector of $A B$ is
$$
\frac{y-4}{x-5}=\frac{1-4}{2-5} \text {. }
$$

Thus, the coordinates $(x, y)$ of the center of the circle passing through $A, B, C$ satisfy
$$
\left\{\begin{array}{l}
y-2=-2(x-4) . \\
y-4=x-5 .
\end{array}\right.
$$

Solving these equations, we get $x=\frac{11}{3}, y=\frac{8}{3}$.