The surface area of the solid generated by the revolution of the circle $(x-6)^{2}+(y-4)^{2}=9$ about its diameter is :

Question

Accepted Answer

To determine the surface area of the solid generated by the revolution of the circle $(x-6)^{2}+(y-4)^{2}=9$ about its diameter, we need to understand the geometry involved.

### Explanation:

1. **Circle Characteristics:**
   - The given equation $(x-6)^{2}+(y-4)^{2}=9$ represents a circle with center at $(6, 4)$ and radius $r = 3$ (since $9 = 3^2$).

2. **Revolution about Diameter:**
   - When a circle is revolved about its diameter, it forms a sphere. The diameter of the circle is $2r = 6$.

3. **Surface Area of the Sphere:**
   - The formula for the surface area of a sphere is $4\pi R^2$, where $R$ is the radius of the sphere.
   - Here, the radius $R$ of the sphere is the same as the radius of the circle, which is $3$.

4. **Calculation:**
   - Substituting $R = 3$ into the formula for the surface area of a sphere, we get:
     $$
     4\pi R^2 = 4\pi (3)^2 = 4\pi \times 9 = 36\pi
     $$

Thus, the surface area of the solid generated by the revolution of the circle about its diameter is $36\pi$, which corresponds to Option 4.

### Why Other Options are Incorrect:

- **Option 1: $4\pi$**
  - This would imply a much smaller sphere, specifically one with a radius of 1, which does not match the given circle's radius of 3.

- **Option 2: $9\pi$**
  - This would suggest a surface area for a sphere with a radius of $\sqrt{9/4} = 1.5$, which is incorrect for the given circle's radius.

- **Option 3: $16\pi$**
  - This would correspond to a sphere with a radius of 2, which again does not match the circle's radius of 3.

In conclusion, the correct surface area of the solid generated by the revolution of the circle about its diameter is indeed $36\pi$, making Option 4 the correct choice.