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### Given:
The equation of the circle \( C \) is:
\[
x^2 + y^2 - 20x - 24y + 195 = 0
\]
The center of the circle is at point \( M \).

### Part (a)
1. **Find the coordinates of point \( M \)** (center of the circle).
   
   To find the center of the circle, we need to rewrite the equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) is the center and \( r \) is the radius.

   We complete the square for \(x\) and \(y\):
   \[
   x^2 - 20x + y^2 - 24y = -195
   \]
   - For \( x^2 - 20x \), complete the square:
     \[
     x^2 - 20x = (x - 10)^2 - 100
     \]
   - For \( y^2 - 24y \), complete the square:
     \[
     y^2 - 24y = (y - 12)^2 - 144
     \]
   
   Now substitute these back into the equation:
   \[
   (x - 10)^2 - 100 + (y - 12)^2 - 144 = -195
   \]
   Simplifying:
   \[
   (x - 10)^2 + (y - 12)^2 = 49
   \]
   
   From this, we can see that the center of the circle \( M \) is at \( (10, 12) \), and the radius is \( \sqrt{49} = 7 \).

2. **The radius of the circle \( C \)** is 7 units.

---

### Part (b)
**Find the length of the line \( MN \).**

The point \( N \) is given as \( (25, 32) \), and \( M \) is \( (10, 12) \).

Using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substitute the values:
\[
MN = \sqrt{(25 - 10)^2 + (32 - 12)^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25
\]
So, the length of the line \( MN \) is 25 units.

---

### Part (c)
**Find the length of the line \( NP \).**

Since the point \( P \) is a point on the circle and the tangent at \( P \) passes through \( N \), the line \( NP \) is the radius of the circle. As calculated earlier, the radius of the circle is 7 units.

Therefore, the length of \( NP \) is 7 units.

---

Would you like a detailed explanation of any step? Here are some additional questions to deepen your understanding:

1. How do you complete the square for a quadratic equation?
2. What is the geometric interpretation of the distance formula used in part (b)?
3. How is the equation of a circle derived in its general form?
4. Why is the length of the line \( NP \) equal to the radius in part (c)?
5. How can the equation of a tangent line to a circle be determined?

**Tip:** When finding the center and radius of a circle from its general equation, completing the square is key to converting it into standard form.

The circle C has equation x^2 + y^2 - 20x - 24y + 195 = 0. The center of C is at the point M. (a) Find (i) the coordinates of the point M, (ii) the radius of the circle C. N is the point with coordinates (25,32). (b) Find the length of the line MN. The tangent to C at a point P on the circle passes through N. (c) Find the length of the line NP.

Learn how to find the center and radius of a circle from its equation and calculate the lengths of line segments MN and NP using the distance formula. This problem involves key concepts in geometry, including tangents to circles. Suitable for students in Grades 10-12.

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