ABC  is a triangle with coordinates A =(2, 6), B =(0, 0), and C =(14, 0).
(a) Let P  be the midpoint of AB. Determine the equation of the line perpendicular to AB passing through P.
(b) Let Q be the point on line BC  for which PQ is perpendicular to AB. Determine the length of AQ. 
(c) There is a (unique) circle passing through the points A, B, and C. Determine the radius of this circle.
Let's solve each part of the problem step-by-step.

### Part (a)
1. **Find the midpoint \( P \) of \( AB \):**
   The coordinates of \( A \) are \( (2, 6) \) and the coordinates of \( B \) are \( (0, 0) \). The midpoint \( P \) of \( AB \) is given by:
   \[
   P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 0}{2}, \frac{6 + 0}{2} \right) = (1, 3)
   \]

2. **Find the slope \( m_{AB} \) of \( AB \):**
   The slope \( m_{AB} \) is given by:
   \[
   m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 6}{0 - 2} = \frac{-6}{-2} = 3
   \]

3. **Find the slope of the line perpendicular to \( AB \):**
   If two lines are perpendicular, the product of their slopes is \( -1 \). Let \( m_\perp \) be the slope of the line perpendicular to \( AB \):
   \[
   m_{AB} \cdot m_\perp = -1 \implies 3 \cdot m_\perp = -1 \implies m_\perp = -\frac{1}{3}
   \]

4. **Find the equation of the line passing through \( P \) with slope \( m_\perp \):**
   Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \):
   \[
   y - 3 = -\frac{1}{3}(x - 1)
   \]
   Simplifying:
   \[
   y - 3 = -\frac{1}{3}x + \frac{1}{3} \implies y = -\frac{1}{3}x + \frac{1}{3} + 3 \implies y = -\frac{1}{3}x + \frac{10}{3}
   \]

### Part (b)
1. **Find the coordinates of \( Q \) on line \( BC \):**
   Since \( Q \) is on line \( BC \) and \( B \) and \( C \) have the same y-coordinate (0), \( Q \) must be of the form \( (q, 0) \).

2. **Use the equation from part (a) to find \( q \):**
   Since \( PQ \) is perpendicular to \( AB \), \( Q \) must lie on the line found in part (a):
   \[
   0 = -\frac{1}{3}q + \frac{10}{3}
   \]
   Solving for \( q \):
   \[
   -\frac{1}{3}q = -\frac{10}{3} \implies q = 10
   \]
   Therefore, \( Q \) is \( (10, 0) \).

3. **Find the length of \( AQ \):**
   Using the distance formula:
   \[
   AQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(10 - 2)^2 + (0 - 6)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10
   \]

### Part (c)
1. **Find the circumcenter of triangle \( ABC \):**
   The circumcenter is the point where the perpendicular bisectors of the sides of the triangle intersect. We already have the perpendicular bisector of \( AB \) from part (a).

2. **Find the perpendicular bisector of \( BC \):**
   The midpoint of \( BC \) is:
   \[
   \left( \frac{0 + 14}{2}, \frac{0 + 0}{2} \right) = (7, 0)
   \]
   Since \( BC \) is horizontal, its perpendicular bisector is vertical and passes through \( (7, 0) \). Thus, the equation is:
   \[
   x = 7
   \]

3. **Find the intersection of the perpendicular bisectors:**
   The intersection of \( y = -\frac{1}{3}x + \frac{10}{3} \) and \( x = 7 \):
   \[
   y = -\frac{1}{3}(7) + \frac{10}{3} = -\frac{7}{3} + \frac{10}{3} = 1
   \]
   So, the circumcenter is \( (7, 1) \).

4. **Find the radius of the circumcircle:**
   The radius is the distance from the circumcenter to any of the vertices. Using \( A(2, 6) \):
   \[
   \text{Radius} = \sqrt{(7 - 2)^2 + (1 - 6)^2} = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
   \]

The final answer is \( \boxed{5\sqrt{2}} \)