26. As shown in Figure $5, A B$ and $C D$ are two perpendicular chords of the same circle, intersecting at point $E$. Given that $A E=12, D E=4$, and $C E=6$. Then the area of the circle is ( ).
(A) $50 \pi$
(B) $45 \pi$
(C) $40 \pi$
(D) $35 \pi$
(E) $30 \pi$
26. A.
$A E \times B E=C E \times D E$, from the given we have $B E=\frac{6 \times 4}{12}=2$.
Take the midpoint $F$ of $A B$, then $O F \perp A B$, take the midpoint $G$ of $D C$, then $O G \perp D C, A F=7, D G=5$, so $O F=G E=1$. Therefore, $O A^{2}=7^{2}+1=50$. Hence, the area of the circle is $50 \pi$.