The radius of the circular section of the sphere `x^(2)+y^(2)+z^(2)=25` by plane `x+y+z=3sqrt3` is

Options

A. 3
B. 4
C. 5
D. none of these

Answer

To find the radius of the circular section of the sphere given by the equation \( x^2 + y^2 + z^2 = 25 \) when intersected by the plane \( x + y + z = 3\sqrt{3} \), we can follow these steps:

### Step 1: Identify the radius of the sphere
The equation of the sphere is given as:
\[ x^2 + y^2 + z^2 = 25 \]
From this equation, we can see that the radius \( R \) of the sphere is:
\[ R = \sqrt{25} = 5 \]

### Step 2: Find the distance from the center of the sphere to the plane
The center of the sphere is at the origin \( O(0, 0, 0) \). The equation of the plane is:
\[ x + y + z = 3\sqrt{3} \]
To find the distance \( d \) from the point \( O(0, 0, 0) \) to the plane, we can use the formula for the distance from a point to a plane:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, \( A = 1, B = 1, C = 1, D = -3\sqrt{3} \), and the coordinates of the point \( (x_1, y_1, z_1) = (0, 0, 0) \).

Substituting these values into the formula gives:
\[ d = \frac{|1(0) + 1(0) + 1(0) - 3\sqrt{3}|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{| - 3\sqrt{3} |}{\sqrt{3}} = \frac{3\sqrt{3}}{\sqrt{3}} = 3 \]

### Step 3: Use the Pythagorean theorem to find the radius of the circular section
Now, we can use the Pythagorean theorem to find the radius \( r \) of the circular section formed by the intersection of the sphere and the plane. According to the theorem:
\[ OP^2 = ON^2 + NP^2 \]
Where:
- \( OP \) is the radius of the sphere (5),
- \( ON \) is the distance from the center of the sphere to the plane (3),
- \( NP \) is the radius of the circular section we want to find.

Substituting the known values:
\[ 5^2 = 3^2 + r^2 \]
\[ 25 = 9 + r^2 \]
\[ r^2 = 25 - 9 = 16 \]
\[ r = \sqrt{16} = 4 \]

### Conclusion
The radius of the circular section of the sphere by the plane is:
\[ \boxed{4} \]