Consider a family of circles passing through the point (3,7) and (6,5). Answer the following questions.
If the circle which belongs to the given family cuts the circle `x^(2)+y^2=29` orthogonally, then the center of that circle is (a) (1 / 2 , 3 / 2 ) (b) ( 9 / 2 , 7 / 2 ) (c) ( 7 / 2 , 9 / 2 ) (d) ( 3 , − 7 / 9 )

Options

A. `(1//2,3//2)`
B. `(9//2,7//2)`
C. `(7//2,9//2)`
D. `(3,-7//9)`

Explanation

To solve the problem, we need to find the center of a circle that passes through the points (3, 7) and (6, 5) and cuts the circle \(x^2 + y^2 = 29\) orthogonally.

### Step-by-step Solution:

1. **Find the Equation of Line AB**:
- The points A(3, 7) and B(6, 5) can be used to find the equation of the line AB.
- The slope \(m\) of line AB is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 7}{6 - 3} = \frac{-2}{3} \]
- Using point-slope form, the equation of the line is:
\[ y - 7 = -\frac{2}{3}(x - 3) \]
- Rearranging gives:
\[ 2x + 3y - 27 = 0 \quad \text{(Equation 1)} \]

**Hint**: Use the slope formula to find the slope of the line connecting two points.

2. **Equation of the Circle with AB as Diameter**:
- The equation of the circle with A and B as endpoints of the diameter is:
\[ (x - 3)(x - 6) + (y - 7)(y - 5) = 0 \]
- Expanding this gives:
\[ (x^2 - 9x + 18) + (y^2 - 12y + 35) = 0 \]
- Simplifying results in:
\[ x^2 + y^2 - 9x - 12y + 53 = 0 \quad \text{(Equation 2)} \]

**Hint**: Remember that the general form of a circle is derived from the product of linear factors.

3. **Orthogonality Condition**:
- The circle \(x^2 + y^2 = 29\) can be rewritten as:
\[ x^2 + y^2 - 29 = 0 \]
- For two circles to intersect orthogonally, the condition is:
\[ 2h_1h_2 + 2k_1k_2 = r_1^2 + r_2^2 \]
- Here, we can find the coefficients from the equations of the circles. For Equation 2, \(h = -\frac{9}{2}\), \(k = -\frac{12}{2}\), and \(r^2 = 53\).

**Hint**: Understand the condition for orthogonality between two circles.

4. **Finding the Value of λ**:
- The family of circles passing through A and B can be expressed as:
\[ (x - 3)(x - 6) + (y - 7)(y - 5) + \lambda(2x + 3y - 27) = 0 \]
- Setting this equal to the orthogonality condition gives:
\[ -29 + 56 + 27\lambda = 0 \]
- Solving for \(\lambda\):
\[ 27\lambda = -27 \implies \lambda = 1 \]

**Hint**: Substitute \(\lambda\) back into the family equation to find the specific circle.

5. **Final Circle Equation**:
- Substitute \(\lambda = 1\) into the family equation:
\[ x^2 + y^2 - 7x - 9y + 26 = 0 \]
- The center of this circle can be found by comparing it to the standard form \((x - h)^2 + (y - k)^2 = r^2\):
\[ h = \frac{7}{2}, \quad k = \frac{9}{2} \]

**Hint**: The center of the circle can be derived from the coefficients of \(x\) and \(y\).

6. **Conclusion**:
- The center of the circle is \(\left(\frac{7}{2}, \frac{9}{2}\right)\).
- Therefore, the correct option is (c) \(\left(\frac{7}{2}, \frac{9}{2}\right)\).