The sum of the square of length of the chord intercepted by the line x+y=n,`ninN` on the circle `x^2+y^2=4`is p then p/11

Options

A. 11
B. 22
C. 33
D. None of these

Answer

To solve the problem, we need to find the sum of the squares of the lengths of the chords intercepted by the line \(x + y = n\) on the circle \(x^2 + y^2 = 4\), where \(n\) is a natural number. We will then compute \(p/11\) where \(p\) is the sum we calculated.

### Step-by-Step Solution:

1. **Identify the Circle and the Line:**
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This is a circle with center at the origin (0, 0) and radius \(r = 2\).

The line is given by:
\[ x + y = n \]
where \(n\) is a natural number.

2. **Find Points of Intersection:**
To find the points where the line intersects the circle, we can substitute \(y = n - x\) into the circle's equation:
\[ x^2 + (n - x)^2 = 4 \]
Expanding this gives:
\[ x^2 + (n^2 - 2nx + x^2) = 4 \]
Simplifying:
\[ 2x^2 - 2nx + n^2 - 4 = 0 \]
Dividing through by 2:
\[ x^2 - nx + \frac{n^2 - 4}{2} = 0 \]

3. **Use the Quadratic Formula:**
The solutions for \(x\) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -n\), and \(c = \frac{n^2 - 4}{2}\):
\[ x = \frac{n \pm \sqrt{n^2 - 2(n^2 - 4)}}{2} \]
Simplifying the discriminant:
\[ n^2 - 2(n^2 - 4) = n^2 - 2n^2 + 8 = 8 - n^2 \]
Thus:
\[ x = \frac{n \pm \sqrt{8 - n^2}}{2} \]

4. **Find Corresponding y-values:**
The corresponding \(y\)-values can be found using \(y = n - x\):
\[ y = n - \frac{n \pm \sqrt{8 - n^2}}{2} = \frac{n \mp \sqrt{8 - n^2}}{2} \]

5. **Calculate Length of the Chord:**
The length of the chord \(L\) can be calculated using the distance formula between the two points of intersection:
\[ L = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
Substituting the values:
\[ L = \sqrt{\left(\frac{\sqrt{8 - n^2}}{2} - \left(-\frac{\sqrt{8 - n^2}}{2}\right)\right)^2 + \left(\frac{\sqrt{8 - n^2}}{2} - \left(-\frac{\sqrt{8 - n^2}}{2}\right)\right)^2} \]
This simplifies to:
\[ L = \sqrt{(2\cdot\frac{\sqrt{8 - n^2}}{2})^2 + (2\cdot\frac{\sqrt{8 - n^2}}{2})^2} = \sqrt{2\cdot(8 - n^2)} = \sqrt{16 - 2n^2} \]

6. **Sum of the Squares of the Lengths:**
Now we need to find the sum of the squares of the lengths for \(n = 1\) and \(n = 2\):
- For \(n = 1\):
\[ L_1^2 = 16 - 2(1^2) = 16 - 2 = 14 \]
- For \(n = 2\):
\[ L_2^2 = 16 - 2(2^2) = 16 - 8 = 8 \]
Thus, the total sum \(p\) is:
\[ p = L_1^2 + L_2^2 = 14 + 8 = 22 \]

7. **Calculate \(p/11\):**
Finally, we calculate:
\[ \frac{p}{11} = \frac{22}{11} = 2 \]

### Final Answer:
\[ \frac{p}{11} = 2 \]