The chords of contact of tangents from three points `A ,Ba n dC` to the circle `x^2+y^2=a^2` are concurrent. Then `A ,B a n dC` will be

Options

A. be concyclic
B. be collinear
C. form the vertices of a triangle
D. none of these

Explanation

To solve the problem, we need to determine the condition under which the chords of contact of tangents from three points \( A, B, \) and \( C \) to the circle \( x^2 + y^2 = a^2 \) are concurrent.

### Step-by-Step Solution:

1. **Identify the Points**:
Let the coordinates of points \( A, B, \) and \( C \) be \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \).

2. **Chords of Contact**:
The equation of the chord of contact from a point \( (x_i, y_i) \) to the circle \( x^2 + y^2 = a^2 \) is given by:
\[ xx_i + yy_i = a^2 \]
Therefore, we have the following equations for points \( A, B, \) and \( C \):
- From \( A \): \( xx_1 + yy_1 = a^2 \) (Equation 1)
- From \( B \): \( xx_2 + yy_2 = a^2 \) (Equation 2)
- From \( C \): \( xx_3 + yy_3 = a^2 \) (Equation 3)

3. **Concurrent Condition**:
For the lines represented by these equations to be concurrent, the determinant of the coefficients must be zero. Thus, we can form the following determinant:
\[ \begin{vmatrix} x_1 & y_1 & a^2 \\ x_2 & y_2 & a^2 \\ x_3 & y_3 & a^2 \end{vmatrix} = 0 \]

4. **Simplifying the Determinant**:
We can simplify this determinant:
\[ \begin{vmatrix} x_1 & y_1 & a^2 \\ x_2 & y_2 & a^2 \\ x_3 & y_3 & a^2 \end{vmatrix} = a^2 \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]
Since \( a^2 \neq 0 \), we have:
\[ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]

5. **Conclusion**:
The determinant being zero implies that the points \( A, B, \) and \( C \) are collinear.

### Final Answer:
Thus, the points \( A, B, \) and \( C \) are **collinear**.

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