The line x `sin alpha - y cos alpha =a ` touches the circle `x ^(2) +y ^(2) =a ^(2),` then,

Options

A. `alpha in [0,pi]`
B. `alpha in [-pi,pi]`
C. `alpha` can have any vlaue
D. `alpha in [-(pi)/(2) , (pi)/(2)]`

Answer

To solve the problem, we need to determine the conditions under which the line \( x \sin \alpha - y \cos \alpha = a \) touches the circle \( x^2 + y^2 = a^2 \).

### Step-by-Step Solution:

1. **Identify the Circle's Properties**:
The equation of the circle is \( x^2 + y^2 = a^2 \). This circle is centered at the origin (0, 0) with a radius \( r = a \).

2. **Equation of the Line**:
The line is given by \( x \sin \alpha - y \cos \alpha = a \). We can rewrite this in the standard form of a line:
\[ \sin \alpha \cdot x - \cos \alpha \cdot y - a = 0 \]
Here, \( A = \sin \alpha \), \( B = -\cos \alpha \), and \( C = -a \).

3. **Calculate the Perpendicular Distance from the Center to the Line**:
The formula for the perpendicular distance \( D \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our case, we need the distance from the center of the circle (0, 0):
\[ D = \frac{|\sin \alpha \cdot 0 - \cos \alpha \cdot 0 - a|}{\sqrt{(\sin \alpha)^2 + (-\cos \alpha)^2}} = \frac{| -a |}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} \]

4. **Simplify the Distance**:
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ D = \frac{a}{1} = a \]

5. **Condition for Tangency**:
For the line to be tangent to the circle, the perpendicular distance \( D \) must equal the radius of the circle. The radius of the circle is also \( a \):
\[ D = a \]
Since we found that \( D = a \), the condition is satisfied.

6. **Conclusion**:
Since the condition holds true for any value of \( \alpha \), we conclude that \( \alpha \) can take any value.

### Final Answer:
The value of \( \alpha \) can be any value.