https://file.aiquickdraw.com/mathbot/file/1726650781509xtq0jm2y.jpg

Let's break down and solve the problem in parts as outlined in the image.

### Part a) Find the equation of line \( AB \)
The points given are \( A(1, 6) \) and \( B(12, 5) \). To find the equation of the line passing through these two points, we first need the slope \( m \), which is calculated by the formula:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Substituting the coordinates of \( A \) and \( B \):

\[
m = \frac{5 - 6}{12 - 1} = \frac{-1}{11}
\]

Now, using the point-slope form of the equation of a line:

\[
y - y_1 = m(x - x_1)
\]

Substituting \( m = -\frac{1}{11} \) and the point \( A(1, 6) \):

\[
y - 6 = -\frac{1}{11}(x - 1)
\]

Simplifying:

\[
y - 6 = -\frac{1}{11}x + \frac{1}{11}
\]
\[
y = -\frac{1}{11}x + \frac{1}{11} + 6
\]
\[
y = -\frac{1}{11}x + \frac{1}{11} + \frac{66}{11}
\]
\[
y = -\frac{1}{11}x + \frac{67}{11}
\]

Thus, the equation of line \( AB \) is:

\[
y = -\frac{1}{11}x + \frac{67}{11}
\]

### Part b) Find the coordinates of point \( R \)
The perpendicular bisector of \( AB \) passes through point \( R \) and \( R \) lies on the x-axis (i.e., \( R = (r, 0) \)).

1. **Midpoint of \( AB \):**
   The midpoint \( M \) of the line segment \( AB \) is given by:

   \[
   M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{1 + 12}{2}, \frac{6 + 5}{2} \right) = (6.5, 5.5)
   \]

2. **Slope of the perpendicular bisector:**
   The slope of the perpendicular bisector is the negative reciprocal of the slope of \( AB \), which is:

   \[
   m_{\text{bisector}} = 11
   \]

   The equation of the perpendicular bisector is in point-slope form, passing through the midpoint \( M(6.5, 5.5) \):

   \[
   y - 5.5 = 11(x - 6.5)
   \]

   Simplifying:

   \[
   y - 5.5 = 11x - 71.5
   \]
   \[
   y = 11x - 66
   \]

3. **Finding \( R \):**
   Since \( R \) lies on the x-axis, \( y = 0 \). Substituting this into the equation of the perpendicular bisector:

   \[
   0 = 11x - 66
   \]
   \[
   x = 6
   \]

   Therefore, \( R = (6, 0) \).

### Part c) Determine the equation of the circle
The equation of a circle is given by:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

Where \( (h, k) \) is the center of the circle and \( r \) is the radius.

- The center of the circle is \( R(6, 0) \).
- The radius is the distance from \( R \) to any point on the circle, such as \( A(1, 6) \).

Using the distance formula to find the radius:

\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(6 - 1)^2 + (0 - 6)^2} = \sqrt{25 + 36} = \sqrt{61}
\]

Thus, the equation of the circle is:

\[
(x - 6)^2 + y^2 = 61
\]

---

Would you like further details on any of these parts, or do you have any specific questions?

Here are some related questions:
1. How do you find the equation of a line given two points?
2. What is the significance of a perpendicular bisector?
3. How do you calculate the distance between two points in a plane?
4. How does the slope of perpendicular lines relate to each other?
5. What is the general form of the equation of a circle?

**Tip:** When finding the equation of a line, always calculate the slope first to determine the angle or steepness of the line.

A circle passes through the points A(1, 6) and B(12, 5) and has a center at point R which lies on the x-axis. 
 a) Find the equation of line AB. 
 b) Given that the perpendicular bisector of AB passes through R, find the coordinates of R. 
 c) Hence or otherwise, determine the equation of the circle.

This problem involves finding the equation of a line and determining the equation of a circle with a given center on the x-axis. The circle passes through points A(1, 6) and B(12, 5), and the solution involves key concepts such as slope, perpendicular bisectors, and distance formulas. Suitable for students in Grades 10-12.

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