Given: \( d + 2 = D \Rightarrow r + 1 = R \)
In the provided diagram, the small circle's radius is denoted by \( r \). Consequently:
Applying the Pythagorean theorem to the right-angled triangle \( \triangle OTB \):
\[ OT^2 + TB^2 = OB^2 \]
Substituting the known values:
\[ r^2 + 3^2 = (r + 1)^2 \]
Expanding the equation:
\[ r^2 + 9 = r^2 + 2r + 1 \]
Simplifying by canceling \( r^2 \) from both sides:
\[ 9 = 2r + 1 \Rightarrow 2r = 8 \Rightarrow r = 4 \]
With \( r = 4 \), the radius of the larger circle is \( R = r + 1 = 5 \).
The diameter of the larger circle is calculated as: \[ D = 2R = 2 \times 5 = \mathbf{10 \, \text{cm}} \]
Correct option: (A) 10 cm