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Let's break down the problem and solve it step by step.

We have:
- Line equation: \( y = 2x + c \)
- Curve equation: \( y = 8 - 2x - x^2 \)

### Part (i) - Finding the value of \( c \) when the line is a tangent to the curve

To find \( c \) when the line is tangent to the curve, the line and the curve must intersect at exactly one point. This means the quadratic equation obtained by setting the two equations equal to each other must have exactly one solution, i.e., a discriminant of 0.

1. **Set the line equation equal to the curve equation:**
   \[
   2x + c = 8 - 2x - x^2
   \]
   Rearrange to form a quadratic equation:
   \[
   x^2 + 4x + (c - 8) = 0
   \]

2. **Condition for tangency:**
   The discriminant (\( \Delta \)) of the quadratic equation must be 0. For a quadratic \( ax^2 + bx + c = 0 \), the discriminant is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   Here, \( a = 1 \), \( b = 4 \), and the constant term is \( c - 8 \). So the discriminant is:
   \[
   \Delta = 4^2 - 4(1)(c - 8) = 16 - 4(c - 8)
   \]
   Simplifying:
   \[
   \Delta = 16 - 4c + 32 = 48 - 4c
   \]

3. **Set the discriminant equal to 0 for tangency:**
   \[
   48 - 4c = 0
   \]
   Solving for \( c \):
   \[
   c = 12
   \]

So, when the line is a tangent to the curve, \( c = 12 \).

---

### Part (ii) - Finding the \( x \)-coordinates of the points of intersection when \( c = 11 \)

Now, we are given that \( c = 11 \). We need to find the points of intersection between the line and the curve by solving the equation:

1. **Set \( y = 2x + 11 \) equal to \( y = 8 - 2x - x^2 \):**
   \[
   2x + 11 = 8 - 2x - x^2
   \]
   Rearrange the equation:
   \[
   x^2 + 4x + 3 = 0
   \]

2. **Solve the quadratic equation:**
   Factor the quadratic equation:
   \[
   (x + 3)(x + 1) = 0
   \]
   So the solutions are:
   \[
   x = -3 \quad \text{or} \quad x = -1
   \]

Thus, the points of intersection are at \( x = -3 \) and \( x = -1 \).

---

### Summary:
- (i) The value of \( c \) for which the line is tangent to the curve is \( c = 12 \).
- (ii) When \( c = 11 \), the \( x \)-coordinates of the points of intersection are \( x = -3 \) and \( x = -1 \).

Do you have any questions or need further explanation?

### Related questions:
1. What happens if the discriminant of a quadratic equation is negative?
2. How do you find the slope of a tangent to a curve at a given point?
3. Can a line intersect a quadratic curve at more than two points? Why or why not?
4. What is the geometric interpretation of the discriminant of a quadratic equation?
5. How can you verify that the line equation is tangent to the curve without solving for \( c \)?

**Tip:** Always check for multiple solutions when dealing with intersections between a line and a curve, as they can intersect at more than one point.

A line has the equation y = 2x + c and a curve has the equation y = 8 - 2x - x^2. (i) For the case where the line is a tangent to the curve, find the value of the constant c. (ii) For the case where c = 11, find the x-coordinates of the points of intersection of the line and the curve.

This problem explores how to find the value of the constant c for which a line is tangent to a quadratic curve, using the discriminant of a quadratic equation. It also involves finding points of intersection for specific values of c. Suitable for high school students in Grades 10-12.

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