Let A(3, 7) and B(6, 5) are two points. `C:x^(2)+y^(2)-4x-6y-3=0` is a circle.
Q. The chords in which the circle C cuts the members of the family S of circle passing through A and B are concurrent at:

Options

A. (2, 3)
B. `(2, (23)/(3))`
C. `(3, (23)/(2))`
D. (3, 2)

Answer

To solve the problem step by step, we will follow the outlined process to find the point of concurrency of the chords formed by the intersection of the given circle and the family of circles passing through points A(3, 7) and B(6, 5).

### Step 1: Write the equation of the family of circles
The family of circles that pass through points A(3, 7) and B(6, 5) can be expressed as:

\[ (x - 3)(x - 6) + (y - 7)(y - 5) + \lambda \begin{vmatrix} x & y & 1 \\ 3 & 7 & 1 \\ 6 & 5 & 1 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
Calculating the determinant:

\[ \begin{vmatrix} x & y & 1 \\ 3 & 7 & 1 \\ 6 & 5 & 1 \end{vmatrix} = x(7 - 5) - y(3 - 6) + 1(3 \cdot 5 - 6 \cdot 7) \]
\[ = 2x + 3y - 27 \]

### Step 3: Substitute the determinant back into the family equation
Substituting the determinant back, we have:

\[ (x - 3)(x - 6) + (y - 7)(y - 5) + \lambda (2x + 3y - 27) = 0 \]

Expanding this gives:

\[ (x^2 - 9x + 18) + (y^2 - 12y + 35) + \lambda (2x + 3y - 27) = 0 \]

### Step 4: Combine and simplify the equation
Combining all terms:

\[ x^2 + y^2 - 9x + 2\lambda x - 12y + 3\lambda y + 53 - 27\lambda = 0 \]

Let this be Equation \( S_2 \).

### Step 5: Write the equation of the given circle
The given circle \( C \) is:

\[ x^2 + y^2 - 4x - 6y - 3 = 0 \]

Let this be Equation \( S_1 \).

### Step 6: Find the common chord
To find the common chord, we subtract \( S_2 \) from \( S_1 \):

\[ S_1 - S_2 = (x^2 + y^2 - 4x - 6y - 3) - (x^2 + y^2 - 9x + 2\lambda x - 12y + 3\lambda y + 53 - 27\lambda) = 0 \]

This simplifies to:

\[ 5x + 6y - 56 + (2\lambda x + 3\lambda y - 27\lambda) = 0 \]

### Step 7: Set up the equations for the chords
Let:

1. \( L_1: 5x + 6y - 56 = 0 \)
2. \( L_2: 2x + 3y - 27 = 0 \)

### Step 8: Solve the equations simultaneously
We can solve these two equations simultaneously.

From \( L_2 \):

\[ 3y = 27 - 2x \implies y = 9 - \frac{2}{3}x \]

Substituting \( y \) in \( L_1 \):

\[ 5x + 6(9 - \frac{2}{3}x) - 56 = 0 \]

This simplifies to:

\[ 5x + 54 - 4x - 56 = 0 \implies x - 2 = 0 \implies x = 2 \]

### Step 9: Substitute back to find \( y \)
Substituting \( x = 2 \) back into \( L_2 \):

\[ 2(2) + 3y - 27 = 0 \implies 4 + 3y - 27 = 0 \implies 3y = 23 \implies y = \frac{23}{3} \]

### Conclusion
The point of concurrency of the chords is:

\[ \boxed{\left(2, \frac{23}{3}\right)} \]