Consider the following diagram, where BC and BD are tangents to the circle, with centre A. The length of each tangent is 4cm and AB=5cm.

To find the radius of the circle, we will use the properties of tangents and the geometry of the circle.

1. **Identify key points and distances**:
- Let \( O \) be the center of the circle (which is denoted as \( A \) in the problem).
- The length of each tangent (BC and BD) is 4 cm.
- The distance from the center \( O \) to a point where the tangent meets the circle (at point \( C \) and \( D \)) is the radius \( r \).

2. **Use the properties of tangents**:
- The distance from the center of the circle \( O \) to the point where the tangent touches the circle is perpendicular to the tangent. Therefore, triangle \( OBC \) (and \( OBD \)) is a right triangle.
- In triangle \( OBC \):
- \( OB \) is the hypotenuse (which represents \( OA \)), and \( OC \) is the radius \( r \).
- The tangent \( BC \) is the other leg of the triangle, measuring 4 cm.

3. **Apply the Pythagorean theorem**:
\[
OB^2 = OC^2 + BC^2
\]
Substituting the known lengths:
\[
AB^2 = r^2 + 4^2
\]
Since \( AB = 5 \) cm,
\[
5^2 = r^2 + 4^2
\]
\[
25 = r^2 + 16
\]

4. **Solve for the radius \( r \)**:
\[
r^2 = 25 - 16
\]
\[
r^2 = 9
\]
\[
r = 3
\]

Thus, the radius of the circle is \( 3 \).