The polars of points (1,7) (2,6) and (t,5) with respect to a circle concurrent then t=

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the value of \( t \) such that the polars of the points \( (1, 7) \), \( (2, 6) \), and \( (t, 5) \) with respect to a circle are concurrent.

### Step-by-Step Solution:

1. **Understanding the Polar Equation**:
The polar of a point \( (x_1, y_1) \) with respect to a circle given by the equation \( x^2 + y^2 = r^2 \) can be expressed as:
\[ x \cdot x_1 + y \cdot y_1 - r^2 = 0 \]
For our case, we can consider the circle to be \( x^2 + y^2 = 1 \) (i.e., \( r^2 = 1 \)).

2. **Finding the Polars**:
- For the point \( (1, 7) \):
\[ 1 \cdot x + 7 \cdot y - 1 = 0 \quad \Rightarrow \quad x + 7y - 1 = 0 \]
- For the point \( (2, 6) \):
\[ 2 \cdot x + 6 \cdot y - 1 = 0 \quad \Rightarrow \quad 2x + 6y - 1 = 0 \]
- For the point \( (t, 5) \):
\[ t \cdot x + 5 \cdot y - 1 = 0 \quad \Rightarrow \quad tx + 5y - 1 = 0 \]

3. **Setting Up the Determinant**:
The three lines are concurrent if the determinant of the coefficients of \( x \), \( y \), and the constant term is zero:
\[ \begin{vmatrix} 1 & 7 & -1 \\ 2 & 6 & -1 \\ t & 5 & -1 \end{vmatrix} = 0 \]

4. **Calculating the Determinant**:
Expanding the determinant:
\[ = 1 \cdot (6 \cdot (-1) - 5 \cdot (-1)) - 7 \cdot (2 \cdot (-1) - t \cdot (-1)) + (-1) \cdot (2 \cdot 5 - 6 \cdot t) \]
Simplifying each term:
\[ = 1 \cdot (-6 + 5) - 7 \cdot (-2 + t) - (10 - 6t) \]
\[ = 1 \cdot (-1) - 7(-2 + t) - (10 - 6t) \]
\[ = -1 + 14 - 7t - 10 + 6t \]
\[ = 3 - t \]

5. **Setting the Determinant to Zero**:
For the lines to be concurrent, we set the determinant to zero:
\[ 3 - t = 0 \]

6. **Solving for \( t \)**:
\[ t = 3 \]

### Final Answer:
Thus, the value of \( t \) is \( \boxed{3} \).