$DEB$ is a chord of a circle such that $DE=3$ and $EB=5$. Let $O$ be the centre of the circle. Join $OE$ and extend $OE$ to cut the circle at $C$. (See diagram). Given $EC=1$, find the radius of the circle.

[asy]
size(6cm);

pair O = (0,0), B = dir(110), D = dir(30), E = 0.4 * B + 0.6 * D, C = intersectionpoint(O--2*E, unitcircle);

draw(unitcircle);
draw(O--C);
draw(B--D);

dot(O);
dot(B);
dot(C);
dot(D);
dot(E);

label("$B$", B, B);
label("$C$", C, C);
label("$D$", D, D);
label("$E$", E, dir(280));
label("$O$", O, dir(270));
[/asy]
1. Let the radius of the circle be \( r \).
2. Since \( O \) is the center of the circle, \( OC \) is a radius, and \( EC = 1 \). Therefore, \( OE = r - 1 \).
3. Extend \( OE \) to cut the circle at \( C \). Since \( O \) is the center, \( OC \) is also a radius, so \( OC = r \).
4. By the Power of a Point theorem (also known as the Chord-Chord Power Theorem), for two intersecting chords \( DE \) and \( EC \) in a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. Therefore, we have:
   \[
   DE \cdot EB = OE \cdot EC
   \]
5. Substituting the given lengths, we get:
   \[
   3 \cdot 5 = (r - 1) \cdot 1
   \]
6. Simplifying the equation:
   \[
   15 = r - 1
   \]
7. Solving for \( r \):
   \[
   r = 16
   \]

The final answer is \(\boxed{16}\).