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. (a)`sqrt(2)`
B. (b)`2`
C. (c)`1`
D. (d)`3`

Answer

To solve the problem of finding 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: Rewrite the equation of the sphere
The equation of the sphere can be rewritten in standard form. We start with:
\[ x^2 + y^2 + z^2 - x + z - 2 = 0 \]
We can rearrange this as:
\[ x^2 - x + y^2 + z^2 + z - 2 = 0 \]
Now, we complete the square for \(x\) and \(z\):

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

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

Substituting these back into the equation gives:
\[ \left(x - \frac{1}{2}\right)^2 + y^2 + \left(z + \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{1}{4} - 2 = 0 \]
This simplifies to:
\[ \left(x - \frac{1}{2}\right)^2 + y^2 + \left(z + \frac{1}{2}\right)^2 = \frac{9}{4} \]
Thus, the center of the sphere is \( \left( \frac{1}{2}, 0, -\frac{1}{2} \right) \) and the radius \( r \) is \( \frac{3}{2} \).

### Step 2: Find the distance from the center of the sphere to the plane
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 our plane \( x + 2y - z - 4 = 0 \), we have \( a = 1, b = 2, c = -1, d = -4 \). The center of the sphere is \( \left( \frac{1}{2}, 0, -\frac{1}{2} \right) \).

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

### Step 3: Find the radius of the circle of intersection
The radius \( R \) of the circle formed by the intersection of the plane and the sphere can be found using the formula:
\[ R = \sqrt{r^2 - d^2} \]
Where \( r \) is the radius of the sphere and \( d \) is the distance from the center of the sphere to the plane.

Substituting the values we found:
\[ R = \sqrt{\left(\frac{3}{2}\right)^2 - \left(\frac{3}{\sqrt{6}}\right)^2} = \sqrt{\frac{9}{4} - \frac{9}{6}} = \sqrt{\frac{27}{12} - \frac{18}{12}} = \sqrt{\frac{9}{12}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Conclusion
The radius of the circle formed by the intersection of the plane and the sphere is \( \frac{\sqrt{3}}{2} \).