A sphere with radius 1 is inscribed in a right circular cone with the radius $r$ of the base.
Let $S$ be the surface area of the cone.

(1) Express $S$ in terms of $r$.

(2) Find the minimum value of $S$.
1. **Express \( S \) in terms of \( r \):**

   Given a sphere with radius 1 inscribed in a right circular cone with base radius \( r \), we need to find the surface area \( S \) of the cone in terms of \( r \).

   Let's denote:
   - \( O \) as the center of the sphere,
   - \( A \) as the point where the sphere touches the cone,
   - \( B \) as the center of the base of the cone,
   - \( C \) as the vertex of the cone,
   - \( D \) as the point where the sphere touches the base of the cone.

   From the problem, we have:
   - \( OB = r \) (radius of the base),
   - \( OA = AD = 1 \) (radius of the sphere),
   - \( OC = h \) (height of the cone).

   Using similar triangles \( \triangle ACD \sim \triangle BCO \):
   \[
   \frac{1}{r} = \frac{h-1}{\sqrt{r^2 + h^2}}
   \]

   Solving for \( h \):
   \[
   \frac{1}{r} = \frac{h-1}{\sqrt{r^2 + h^2}}
   \]
   \[
   \sqrt{r^2 + h^2} = r(h-1)
   \]
   Squaring both sides:
   \[
   r^2 + h^2 = r^2(h-1)^2
   \]
   \[
   r^2 + h^2 = r^2(h^2 - 2h + 1)
   \]
   \[
   r^2 + h^2 = r^2h^2 - 2r^2h + r^2
   \]
   \[
   h^2 - r^2h^2 = -2r^2h
   \]
   \[
   h^2(1 - r^2) = -2r^2h
   \]
   \[
   h = \frac{2r^2}{r^2 - 1}
   \]

   Next, we find the slant height \( BC \) of the cone:
   \[
   BC = \sqrt{r^2 + h^2} = \sqrt{r^2 + \left(\frac{2r^2}{r^2 - 1}\right)^2}
   \]
   \[
   BC = \sqrt{r^2 + \frac{4r^4}{(r^2 - 1)^2}}
   \]
   \[
   BC = \frac{r(r^2 + 1)}{r^2 - 1}
   \]

   The surface area \( S \) of the cone is the sum of the lateral surface area and the base area:
   \[
   S(r) = \pi r \cdot BC + \pi r^2
   \]
   \[
   S(r) = \pi r \cdot \frac{r(r^2 + 1)}{r^2 - 1} + \pi r^2
   \]
   \[
   S(r) = \pi \frac{r^2(r^2 + 1)}{r^2 - 1} + \pi r^2
   \]
   \[
   S(r) = \pi \frac{r^4 + r^2}{r^2 - 1} + \pi r^2
   \]
   \[
   S(r) = \pi \frac{r^4 + r^2 + r^2 - r^2}{r^2 - 1}
   \]
   \[
   S(r) = \pi \frac{r^4 + 2r^2}{r^2 - 1}
   \]
   \[
   S(r) = \pi \frac{2r^4}{r^2 - 1}
   \]

2. **Find the minimum value of \( S \):**

   To find the minimum value of \( S \), we need to find the critical points by taking the derivative of \( S(r) \) and setting it to zero:
   \[
   S'(r) = \frac{d}{dr} \left( \pi \frac{2r^4}{r^2 - 1} \right)
   \]
   Using the quotient rule:
   \[
   S'(r) = \pi \left( \frac{(8r^3)(r^2 - 1) - 2r^4(2r)}{(r^2 - 1)^2} \right)
   \]
   \[
   S'(r) = \pi \left( \frac{8r^5 - 8r^3 - 4r^5}{(r^2 - 1)^2} \right)
   \]
   \[
   S'(r) = \pi \left( \frac{4r^5 - 8r^3}{(r^2 - 1)^2} \right)
   \]
   \[
   S'(r) = \pi \left( \frac{4r^3(r^2 - 2)}{(r^2 - 1)^2} \right)
   \]
   Setting \( S'(r) = 0 \):
   \[
   4r^3(r^2 - 2) = 0
   \]
   \[
   r^3 = 0 \quad \text{or} \quad r^2 - 2 = 0
   \]
   \[
   r = 0 \quad \text{or} \quad r = \sqrt{2}
   \]

   Since \( r = 0 \) is not a valid radius for the cone, we have:
   \[
   r = \sqrt{2}
   \]

   Substituting \( r = \sqrt{2} \) into \( S(r) \):
   \[
   S(\sqrt{2}) = \pi \frac{2(\sqrt{2})^4}{(\sqrt{2})^2 - 1}
   \]
   \[
   S(\sqrt{2}) = \pi \frac{2 \cdot 4}{2 - 1}
   \]
   \[
   S(\sqrt{2}) = \pi \frac{8}{1}
   \]
   \[
   S(\sqrt{2}) = 8\pi
   \]

The final answer is \( \boxed{ 8\pi } \).