For all values of `m in R` the line `y - mx + m - 1 = 0` cuts the circle `x^2 + y^2 - 2x - 2y + 1 = 0` at an angle

Options

A. `(pi)/(3)`
B. `(pi)/(6)`
C. `(pi)/(2)`
D. `(pi)/(4)`

Explanation

To solve the problem, we need to determine the angle at which the line \( y - mx + m - 1 = 0 \) intersects the circle defined by the equation \( x^2 + y^2 - 2x - 2y + 1 = 0 \).

### Step 1: Rewrite the equation of the circle in standard form.

The given equation of the circle is:

\[ x^2 + y^2 - 2x - 2y + 1 = 0 \]

To convert this into standard form, we complete the square for both \( x \) and \( y \).

1. For \( x^2 - 2x \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

2. For \( y^2 - 2y \):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 + 1 = 0 \]

Simplifying this results in:

\[ (x - 1)^2 + (y - 1)^2 - 1 = 0 \]

Thus, we can rewrite it as:

\[ (x - 1)^2 + (y - 1)^2 = 1 \]

This shows that the circle has a center at \( (1, 1) \) and a radius of \( 1 \).

### Step 2: Determine if the line passes through the center of the circle.

The line is given by:

\[ y - mx + m - 1 = 0 \]

To check if this line passes through the center of the circle \( (1, 1) \), we substitute \( x = 1 \) and \( y = 1 \) into the line equation:

\[ 1 - m(1) + m - 1 = 0 \]

This simplifies to:

\[ 1 - m + m - 1 = 0 \]

This equation holds true, confirming that the line passes through the center of the circle.

### Step 3: Determine the angle at which the line intersects the circle.

Since the line passes through the center of the circle, it acts as a diameter of the circle. The tangent to the circle at any point is perpendicular to the radius at that point. Therefore, the angle between the line (which is the diameter) and the tangent at the point of intersection is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

### Conclusion

Thus, the angle at which the line cuts the circle is:

\[ \frac{\pi}{2} \]