The plane `x+2y-z=4` cuts the sphere `x^(2)+y^(2)+z^(2)-x+z-2=0` in a circle of radius

Options

A. `sqrt(2)`
B. `2`
C. `1`
D. `3`

Answer

To find the radius of the circle formed by the intersection of the plane \( x + 2y - z = 4 \) and the sphere \( x^2 + y^2 + z^2 - x + z - 2 = 0 \), we will follow these steps:

### Step 1: Identify the center and radius of the sphere
The equation of the sphere can be rewritten in standard form. The general equation of a sphere is given by:

\[ (x - a)^2 + (y - b)^2 + (z - c)^2 = R^2 \]

To convert the given sphere equation \( x^2 + y^2 + z^2 - x + z - 2 = 0 \) into this form, we complete the square for each variable.

1. **For \( x \)**:
\[ x^2 - x = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} \]

2. **For \( y \)**:
\[ y^2 = (y - 0)^2 \]

3. **For \( z \)**:
\[ z^2 + z = \left(z + \frac{1}{2}\right)^2 - \frac{1}{4} \]

Now substituting these back into the sphere equation:

\[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + (y - 0)^2 + \left(z + \frac{1}{2}\right)^2 - \frac{1}{4} - 2 = 0 \]

Combining the constant terms:

\[ \left(x - \frac{1}{2}\right)^2 + (y - 0)^2 + \left(z + \frac{1}{2}\right)^2 - \frac{1}{2} - 2 = 0 \]

This simplifies to:

\[ \left(x - \frac{1}{2}\right)^2 + (y - 0)^2 + \left(z + \frac{1}{2}\right)^2 = \frac{5}{2} \]

Thus, the center of the sphere is \( C\left(\frac{1}{2}, 0, -\frac{1}{2}\right) \) and the radius \( R \) is \( \sqrt{\frac{5}{2}} \).

### Step 2: Find the perpendicular distance from the center of the sphere to the plane
The formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For the plane \( x + 2y - z = 4 \), we rewrite it as \( x + 2y - z - 4 = 0 \), where \( A = 1, B = 2, C = -1, D = -4 \).

Substituting the center \( C\left(\frac{1}{2}, 0, -\frac{1}{2}\right) \):

\[ d = \frac{\left| 1 \cdot \frac{1}{2} + 2 \cdot 0 - 1 \cdot \left(-\frac{1}{2}\right) - 4 \right|}{\sqrt{1^2 + 2^2 + (-1)^2}} = \frac{\left| \frac{1}{2} + \frac{1}{2} - 4 \right|}{\sqrt{6}} = \frac{\left| 1 - 4 \right|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2} \]

### Step 3: Use the relationship between the radius of the sphere, the radius of the circle, and the distance from the center to the plane
The relationship is given by:

\[ R^2 = r^2 + d^2 \]

Where:
- \( R = \sqrt{\frac{5}{2}} \)
- \( d = \frac{3}{\sqrt{6}} \)

Calculating \( R^2 \):

\[ R^2 = \frac{5}{2} \]

Calculating \( d^2 \):

\[ d^2 = \left(\frac{3}{\sqrt{6}}\right)^2 = \frac{9}{6} = \frac{3}{2} \]

Now substituting into the equation:

\[ \frac{5}{2} = r^2 + \frac{3}{2} \]

Solving for \( r^2 \):

\[ r^2 = \frac{5}{2} - \frac{3}{2} = 1 \]

Thus, the radius \( r \) is:

\[ r = \sqrt{1} = 1 \]

### Final Answer
The radius of the circle formed by the intersection of the plane and the sphere is \( 1 \).

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