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Let's go through the problem step by step. Here's a breakdown of each part:

**Given equation of the circle:**
\[
(x - 2)^2 + \left( y + 1 \right)^2 = \frac{169}{4}
\]

This is the standard equation of a circle:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Where:
- \((h, k)\) is the center of the circle,
- \(r\) is the radius.

### (a) Write down the coordinates of \(N\).
From the equation, the center \(N\) is at:
\[
N = (h, k) = (2, -1)
\]

### (b) Find the radius of \(C\).
The right-hand side of the equation represents \(r^2 = \frac{169}{4}\).
So the radius \(r\) is:
\[
r = \sqrt{\frac{169}{4}} = \frac{13}{2} = 6.5
\]

### (c) Find the coordinates of \(A\) and \(B\).

- We are told that the chord \(AB\) is parallel to the \(x\)-axis and is 12 units long. This means the points \(A\) and \(B\) lie horizontally across from each other on the circle.

- The center of the circle is at \(N(2, -1)\), and since \(A\) and \(B\) are symmetric about the center and parallel to the \(x\)-axis, the \(y\)-coordinates of both \(A\) and \(B\) will be the same as the center's \(y\)-coordinate, i.e., \(-1\).

- The distance \(AB = 12\), so the points \(A\) and \(B\) will be 6 units left and right from the center \(N(2, -1)\), along the \(x\)-axis. Therefore:
  - The \(x\)-coordinate of \(A\) is \(2 - 6 = -4\),
  - The \(x\)-coordinate of \(B\) is \(2 + 6 = 8\).

Thus:
\[
A = (-4, -1), \quad B = (8, -1)
\]

### (d) Show that \(\angle ANB = 134.8^\circ\).

To find the angle \(\angle ANB\), we can use the cosine rule in the triangle \(ANB\).

The length of the radius is \(6.5\) units, so both \(NA\) and \(NB\) are equal to 6.5.

Using the cosine rule for angle \(\theta = \angle ANB\):
\[
\cos(\theta) = \frac{NA^2 + NB^2 - AB^2}{2 \cdot NA \cdot NB}
\]
Substitute the values:
\[
NA = NB = 6.5, \quad AB = 12
\]
\[
\cos(\theta) = \frac{6.5^2 + 6.5^2 - 12^2}{2 \cdot 6.5 \cdot 6.5}
\]
\[
\cos(\theta) = \frac{42.25 + 42.25 - 144}{84.5}
\]
\[
\cos(\theta) = \frac{84.5 - 144}{84.5} = \frac{-59.5}{84.5}
\]
\[
\cos(\theta) \approx -0.704
\]
\[
\theta = \cos^{-1}(-0.704) \approx 134.8^\circ
\]

### (e) Find the length \(AP\).

The point \(P\) is where the tangents to the circle at points \(A\) and \(B\) meet. Since the tangents to a circle from a common external point are equal, \(AP = BP\).

We can use the tangent-secant theorem to find the length of \(AP\):
\[
AP^2 = OA^2 - r^2
\]
where \(OA\) is the distance from \(O\) (the origin) to \(A\), and \(r\) is the radius of the circle.

However, we'll need to clarify how to approach this based on the actual problem setup.

---

Do you want details on how to approach part (e), or any other questions?

### Related Questions:
1. How is the cosine rule derived for triangles in general?
2. Can you explain the properties of tangents to circles in more detail?
3. What other methods can we use to find angles in triangles?
4. How do we derive the standard form of a circle’s equation from its definition?
5. Can you demonstrate the steps for calculating arc lengths and sectors in circles?

**Tip:** Always sketch the geometry and label known quantities to keep track of relationships easily.

The figure shows a sketch of a circle C with center N, and the equation (x - 2)^2 + (y + 1)^2 = 169/4. The problem involves finding coordinates, calculating angles, and applying geometry and algebra.

This problem involves finding the coordinates of a circle’s center and radius, using the chord properties, and applying the Cosine Rule to calculate angles. The problem is suitable for students in Grades 10-12, covering key concepts in coordinate geometry and circle theorems.

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