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Let's break down question 11c, which asks to find the length of the line \( NP \).

### Given:
- The equation of the circle: 
  \[
  x^2 + y^2 - 20x - 24y + 195 = 0
  \]
- Point \( N \) has coordinates \( (25, 32) \).
- The tangent to the circle at point \( P \) passes through point \( N \).

### Step 1: Rewrite the equation of the circle in standard form
We need to rewrite the equation of the circle in the form:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Where \( (h, k) \) is the center and \( r \) is the radius.

Start by completing the square:

1. Group the \( x \)-terms and \( y \)-terms:
   \[
   (x^2 - 20x) + (y^2 - 24y) = -195
   \]

2. Complete the square for both \( x \) and \( y \):
   - For \( x^2 - 20x \): take half of \( -20 \), which is \( -10 \), then square it to get \( 100 \). Add and subtract \( 100 \).
   - For \( y^2 - 24y \): take half of \( -24 \), which is \( -12 \), then square it to get \( 144 \). Add and subtract \( 144 \).

   Now the equation becomes:
   \[
   (x^2 - 20x + 100) + (y^2 - 24y + 144) = -195 + 100 + 144
   \]
   Simplifying the right-hand side:
   \[
   (x - 10)^2 + (y - 12)^2 = 49
   \]

Thus, the circle is centered at \( M(10, 12) \) and has a radius of \( r = 7 \) (since \( 49 = 7^2 \)).

### Step 2: Find the length of \( NP \)
Since point \( P \) lies on the circle and the tangent at \( P \) passes through point \( N(25, 32) \), the length of the tangent from a point to a circle can be calculated using the formula:
\[
\text{Length of tangent} = \sqrt{d^2 - r^2}
\]
Where:
- \( d \) is the distance from point \( N(25, 32) \) to the center of the circle \( M(10, 12) \),
- \( r \) is the radius of the circle (which we found to be \( 7 \)).

1. First, find the distance \( d \) between \( N(25, 32) \) and \( M(10, 12) \) using the distance formula:
   \[
   d = \sqrt{(25 - 10)^2 + (32 - 12)^2}
   \]
   \[
   d = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25
   \]

2. Now, use the tangent length formula:
   \[
   \text{Length of } NP = \sqrt{d^2 - r^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24
   \]

Thus, the length of \( NP \) is **24 units**.

---

Would you like further clarification or more details on any step?

Here are 5 related questions you might find helpful:
1. How do you find the equation of the tangent line to a circle at a specific point?
2. What is the general method for completing the square on a quadratic equation?
3. How can you derive the formula for the length of a tangent from a point to a circle?
4. What are the properties of tangent lines to circles in different geometrical contexts?
5. How can the Pythagorean theorem be applied in other circle-related problems?

**Tip**: When completing the square, remember to balance the equation by adding the same values to both sides!

Find the length of the line NP where the circle has the equation x^2 + y^2 - 20x - 24y + 195 = 0, and the tangent to the circle at point P passes through point N(25, 32).

Learn how to find the length of the tangent NP to a circle given the equation x^2 + y^2 - 20x - 24y + 195 = 0. This problem uses coordinate geometry, completing the square, and the tangent length formula. Ideal for Grades 10-12 students.

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