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Let's solve the problem step by step.

Given:
- Points \( C = (6, p) \)
- \( D = (2, p + 2) \)
- \( E = (-1, p + 1) \)
- Center of the circle is \( (p, q) \)

### Part I: Equation of the perpendicular bisector of \( CD \)

To find the perpendicular bisector of line segment \( CD \):
1. **Find the midpoint of \( CD \):**
   \[
   \text{Midpoint} = \left( \frac{6 + 2}{2}, \frac{p + (p + 2)}{2} \right) = (4, p + 1)
   \]

2. **Find the slope of \( CD \):**
   \[
   \text{Slope of } CD = \frac{(p + 2) - p}{2 - 6} = \frac{2}{-4} = -\frac{1}{2}
   \]

3. **Find the slope of the perpendicular bisector:**
   The slope of the perpendicular bisector is the negative reciprocal:
   \[
   \text{Slope of perpendicular bisector} = 2
   \]

4. **Equation of the perpendicular bisector using point-slope form:**
   Using the midpoint \((4, p + 1)\) and slope \(2\):
   \[
   y - (p + 1) = 2(x - 4)
   \]
   Simplify this equation:
   \[
   y - p - 1 = 2x - 8
   \]
   \[
   y = 2x - 7 + p
   \]
   Rearranging:
   \[
   3p + q = 7
   \]

### Part II: Equation of the perpendicular bisector of \( DE \)

To find the perpendicular bisector of line segment \( DE \):
1. **Find the midpoint of \( DE \):**
   \[
   \text{Midpoint} = \left( \frac{2 + (-1)}{2}, \frac{(p + 2) + (p + 1)}{2} \right) = \left( \frac{1}{2}, p + \frac{3}{2} \right)
   \]

2. **Find the slope of \( DE \):**
   \[
   \text{Slope of } DE = \frac{(p + 1) - (p + 2)}{-1 - 2} = \frac{-1}{-3} = \frac{1}{3}
   \]

3. **Find the slope of the perpendicular bisector:**
   The slope of the perpendicular bisector is the negative reciprocal:
   \[
   \text{Slope of perpendicular bisector} = -3
   \]

4. **Equation of the perpendicular bisector using point-slope form:**
   Using the midpoint \(\left( \frac{1}{2}, p + \frac{3}{2} \right)\) and slope \(-3\):
   \[
   y - \left( p + \frac{3}{2} \right) = -3 \left( x - \frac{1}{2} \right)
   \]
   Simplify this equation:
   \[
   y - p - \frac{3}{2} = -3x + \frac{3}{2}
   \]
   Rearrange to standard form:
   \[
   2p + q = 3
   \]

### Part III: Solving for the values of \( p \) and \( q \)

We now have the system of equations:
1. \( 3p + q = 7 \)
2. \( 2p + q = 3 \)

Subtract the second equation from the first:
\[
(3p + q) - (2p + q) = 7 - 3
\]
\[
p = 4
\]

Substitute \( p = 4 \) into the second equation:
\[
2(4) + q = 3
\]
\[
8 + q = 3
\]
\[
q = -5
\]

### Final Answer
The values are:
- \( p = 4 \)
- \( q = -5 \)

Would you like more details or further clarifications on this solution?

Here are some related questions for further exploration:
1. How do you find the midpoint of a line segment in coordinate geometry?
2. What is the method to determine the slope of a line given two points?
3. Why is the perpendicular bisector important in circle geometry?
4. How can you solve a system of linear equations using substitution?
5. What are other methods to find the center of a circle given its points?

**Tip:** The slope of the perpendicular bisector is always the negative reciprocal of the original slope, which helps in quickly determining its equation.

The points C, D, and E have coordinates (6, p), (2, p + 2), and (-1, p + 1), respectively, and lie on a circle with center (p, q). Find the perpendicular bisectors of CD and DE, and deduce the values of p and q.

Learn how to find the perpendicular bisectors of chords in a circle, specifically for the points C(6, p), D(2, p + 2), and E(-1, p + 1), and solve for the values of p and q. Includes detailed steps on finding midpoints, slopes, and solving systems of equations.

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