Consider a family of circles passing through the point (3,7) and (6,5). Answer the following questions.
If each circle in the family cuts the circle `x^(2)+y^(2)-4x-6y-3=0`, then all the common chords pass through the fixed point which is

Options

A. `(1,23)`
B. `(2, 23//2)`
C. `(-3,3//2)`
D. none of these

Explanation

To solve the problem, we need to find the fixed point through which all common chords of the family of circles passing through the points (3,7) and (6,5) intersect, given that these circles cut the circle defined by the equation \(x^2 + y^2 - 4x - 6y - 3 = 0\).

### Step 1: Rewrite the given circle equation
We start by rewriting the equation of the circle in standard form. The given equation is:

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

We can complete the square for the \(x\) and \(y\) terms.

For \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

For \(y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

Substituting these back into the equation gives:

\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 - 3 = 0 \]

Simplifying this:

\[ (x - 2)^2 + (y - 3)^2 - 16 = 0 \]

Thus, we have:

\[ (x - 2)^2 + (y - 3)^2 = 16 \]

This represents a circle centered at (2, 3) with a radius of 4.

### Step 2: Determine the family of circles
The family of circles passing through the points (3,7) and (6,5) can be expressed in the general form:

\[ (x - 3)(x - h) + (y - 7)(y - k) = 0 \]

Where (h, k) is the center of the circle. Since the circles pass through (3, 7) and (6, 5), we can derive the equation of the chord joining these points.

### Step 3: Find the equation of the chord
The midpoint of the points (3, 7) and (6, 5) is:

\[ \left(\frac{3 + 6}{2}, \frac{7 + 5}{2}\right) = \left(\frac{9}{2}, 6\right) \]

The slope of the line joining (3, 7) and (6, 5) is:

\[ \text{slope} = \frac{5 - 7}{6 - 3} = \frac{-2}{3} \]

The equation of the line (chord) can be written in point-slope form as:

\[ y - 7 = -\frac{2}{3}(x - 3) \]

Simplifying this gives:

\[ y = -\frac{2}{3}x + 7 + 2 = -\frac{2}{3}x + 9 \]

### Step 4: Find the fixed point of the common chords
To find the fixed point where all common chords intersect, we can use the property that the radical axis of the two circles will pass through this point. The radical axis is perpendicular to the line joining the centers of the two circles.

Since the family of circles passes through (3, 7) and (6, 5), we can find the intersection of the line \(y = -\frac{2}{3}x + 9\) with the center of the circle (2, 3).

Substituting \(x = 2\) into the line equation:

\[ y = -\frac{2}{3}(2) + 9 = -\frac{4}{3} + 9 = \frac{23}{3} \]

Thus, the fixed point through which all common chords pass is:

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

### Final Answer
The fixed point through which all common chords pass is \(\left(2, \frac{23}{3}\right)\).