If the line ` y cos alpha = x sin alpha +a cos alpha ` be a tangent to the circle `x^(2)+y^(2)=a^(2)`, then

Options

A. `sin^(2)alpha=1`
B. `cos^(2) alpha=1`
C. `sin ^(2)alpha = a^(2)`
D. `cos^(2) alpha=a^(2)`

Explanation

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

### Step-by-Step Solution:

1. **Rewrite the line equation**:
The given line can be rearranged to the standard form:
\[ x \sin \alpha - y \cos \alpha + a \cos \alpha = 0 \]

2. **Identify the circle's properties**:
The equation of the circle is \( x^2 + y^2 = a^2 \). The center of the circle is at \( (0, 0) \) and the radius \( r \) is \( a \).

3. **Use the distance formula from a point to a line**:
The distance \( d \) from the center of the circle \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = \sin \alpha \), \( B = -\cos \alpha \), and \( C = a \cos \alpha \). Thus, substituting \( (x_1, y_1) = (0, 0) \):
\[ d = \frac{|\sin \alpha \cdot 0 - \cos \alpha \cdot 0 + a \cos \alpha|}{\sqrt{(\sin \alpha)^2 + (-\cos \alpha)^2}} = \frac{|a \cos \alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} \]

4. **Simplify the distance**:
Since \( \sin^2 \alpha + \cos^2 \alpha = 1 \):
\[ d = |a \cos \alpha| \]

5. **Set the distance equal to the radius**:
For the line to be tangent to the circle, the distance \( d \) must equal the radius \( a \):
\[ |a \cos \alpha| = a \]

6. **Solve for the condition**:
This gives us two cases:
- Case 1: \( a \cos \alpha = a \) which simplifies to \( \cos \alpha = 1 \) (i.e., \( \alpha = 0 \)).
- Case 2: \( a \cos \alpha = -a \) which simplifies to \( \cos \alpha = -1 \) (i.e., \( \alpha = \pi \)).
Thus, the condition for the line to be tangent to the circle is:
\[ \cos^2 \alpha = 1 \]

### Final Condition:
The required condition is:
\[ \cos^2 \alpha = 1 \]