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

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

Options

A. (a)160
B. (b)320
C. (c)105
D. (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 Line:**
The equation of the circle is given by:
\[ x^2 + y^2 = 16 \]
This represents a circle with a radius of \(4\) (since \(r^2 = 16\)) centered at the origin \((0,0)\).

The equation of the line is:
\[ x + y = n \]
We can rewrite this as:
\[ x + y - n = 0 \]

2. **Find the Perpendicular Distance from the Origin to the Line:**
The formula for the distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(x + y - n = 0\) (where \(A = 1\), \(B = 1\), and \(C = -n\)), the distance from the origin \((0, 0)\) is:
\[ d = \frac{|1 \cdot 0 + 1 \cdot 0 - n|}{\sqrt{1^2 + 1^2}} = \frac{n}{\sqrt{2}} \]

3. **Use Pythagorean Theorem:**
In the right triangle formed by the radius \(OA\), the perpendicular distance \(ON\), and half the chord length \(AN\):
\[ OA^2 = ON^2 + AN^2 \]
Here, \(OA = 4\) (the radius), \(ON = \frac{n}{\sqrt{2}}\), and \(AN\) is the half-length of the chord.

Thus, we have:
\[ 16 = \left(\frac{n}{\sqrt{2}}\right)^2 + AN^2 \]
Simplifying gives:
\[ 16 = \frac{n^2}{2} + AN^2 \]
Rearranging for \(AN^2\):
\[ AN^2 = 16 - \frac{n^2}{2} \]

4. **Find the Length of the Chord:**
The length of the chord \(AB\) is:
\[ AB = 2 \cdot AN \]
Therefore, the square of the length of the chord is:
\[ AB^2 = 4 \cdot AN^2 = 4 \left(16 - \frac{n^2}{2}\right) = 64 - 2n^2 \]

5. **Sum the Squares of the Chord Lengths for \(n = 1\) to \(5\):**
We need to calculate:
\[ \sum_{n=1}^{5} AB^2 = \sum_{n=1}^{5} (64 - 2n^2) \]
This can be split into two separate sums:
\[ = \sum_{n=1}^{5} 64 - \sum_{n=1}^{5} 2n^2 \]
The first sum is:
\[ 64 \cdot 5 = 320 \]
The second sum involves 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,
\[ 2 \sum_{n=1}^{5} n^2 = 2 \cdot 55 = 110 \]

Therefore, the total sum is:
\[ 320 - 110 = 210 \]

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