The sum of the squares of the length of the chords intercepted on the circle `x^(2)+y^(2)=16`, by the lines x +y = n , `n in N`, where N is the set of all natural numbers, is

Options

A. 320
B. 105
C. 160
D. 210

Explanation

To solve the problem, we need to find the sum of the squares of the lengths of the chords intercepted on the circle \( x^2 + y^2 = 16 \) by the lines \( x + y = n \), where \( n \) is a natural number.

### Step-by-Step Solution:

1. **Identify the Circle and its Properties**:
The equation of the circle is \( x^2 + y^2 = 16 \). This means the radius \( r \) of the circle is \( \sqrt{16} = 4 \).

2. **Identify the Line**:
The line is given by \( x + y = n \). We can rewrite it in standard form as \( x + y - n = 0 \).

3. **Calculate the Perpendicular Distance from the Center to the Line**:
The center of the circle is at the origin \( (0, 0) \). The formula for the distance \( D \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = 1, B = 1, C = -n \), and \( (x_1, y_1) = (0, 0) \):
\[ D = \frac{|1 \cdot 0 + 1 \cdot 0 - n|}{\sqrt{1^2 + 1^2}} = \frac{| - n |}{\sqrt{2}} = \frac{n}{\sqrt{2}} \]

4. **Relate the Length of the Chord to the Perpendicular Distance**:
The length of the chord \( AB \) can be derived from the right triangle formed by the radius, the perpendicular distance, and half the chord length. Using the Pythagorean theorem:
\[ OA^2 = ON^2 + AN^2 \]
where \( OA = 4 \) (radius), \( ON = \frac{n}{\sqrt{2}} \), and \( AN \) is half the length of the chord. Thus:
\[ 4^2 = \left(\frac{n}{\sqrt{2}}\right)^2 + AN^2 \]
\[ 16 = \frac{n^2}{2} + AN^2 \]
Rearranging gives:
\[ AN^2 = 16 - \frac{n^2}{2} \]

5. **Calculate the Length of the Chord**:
Since \( AB = 2 \cdot AN \):
\[ AB^2 = (2 \cdot AN)^2 = 4 \cdot AN^2 = 4 \left(16 - \frac{n^2}{2}\right) = 64 - 2n^2 \]

6. **Sum the Squares of the Lengths of the Chords for Natural Numbers \( n \)**:
We need to sum \( AB^2 \) for \( n = 1, 2, 3, 4, 5 \):
\[ S = \sum_{n=1}^{5} (64 - 2n^2) = \sum_{n=1}^{5} 64 - \sum_{n=1}^{5} 2n^2 \]
The first term:
\[ \sum_{n=1}^{5} 64 = 5 \cdot 64 = 320 \]
The second term (using the formula for the sum of squares):
\[ \sum_{n=1}^{5} n^2 = \frac{5(5 + 1)(2 \cdot 5 + 1)}{6} = \frac{5 \cdot 6 \cdot 11}{6} = 55 \]
Thus:
\[ S = 320 - 2 \cdot 55 = 320 - 110 = 210 \]

### Final Answer:
The sum of the squares of the lengths of the chords is \( \boxed{210} \).