Question: Problem : 

The sum of the squares of the length of the chord intercepted by the line x+y=n $n \in N$ on the circle $x^2+y^2=4$ is 

(a) 11 

(b) 22 

(c) 33 

(d) 13

I am unable to understand this problem request you to please ellaborate on this , I will be greatful to you thanks...

Answer: 
Let us find the intersections :

$$x^2+(n-x)^2=4\iff 2x^2-2nx+n^2-4=0\ \ \ \ (1)$$

If $\displaystyle x\_1,x\_2$ are the abscissa of the intersections, $\displaystyle x\_1+x\_2=\frac{2n}2=n$ and $\displaystyle x\_1\cdot x\_2=\frac{n^2-4}2$

So, the length of the chord for a particular $n$ will be $$\sqrt{(x\_1-x\_2)^2+(y\_1-y\_2)^2}=\sqrt{(x\_1-x\_2)^2+\{n-x\_1-(n-x\_2)\}^2}=\sqrt2\sqrt{(x\_1-x\_2)^2}$$

Now, $\displaystyle(x\_1-x\_2)^2=(x\_1+x\_2)^2-4x\_1x\_2=n^2-2(n^2-4)=8-n^2$ which needs to be $\ge0$ for the real intersection.

Also using discriminant on $(1),$ for the real intersection, $$(-2n)^2-4\cdot2(n^2-4)\ge0\iff n^2\le8\iff-2\sqrt2\le n\le2\sqrt2$$

Hope you can take it home from here?
Referring to the diagram below:-

![enter image description here](https://i.stack.imgur.com/wj1un.png)

First of all, we have to determine the allowable values of n and found that n can only be -2, -1, 0, 1 or 2 (left as an exercise).

The cases when n= 2 or -2, $L^2$ is 8.

The case when n= 0, $L^2 = (2\* radius)^2 = 16$

The case when n = 1, the points of intersection are $(\frac {\sqrt 7 - 1}{2}, 1 - \frac {\sqrt 7 - 1}{2})$ and $(\frac {-\sqrt 7 - 1}{2}, 1 - \frac {-\sqrt 7 - 1}{2})$

Then, the corresponding $L^2$ = … by distance formula … = 14

Similarly, the corresponding $L^2$ for n = -1 is also 14 by symmetry.

The required sum is therefore …. = 60 = ***none of the given options***.

Even if the case of n = 0 is not counted, the answer will then be 44 = ***none of the given options***. 

Note also that this question should be treated by a more advanced level method (by consider sum and product of roots of a quadratic equation.