The sum of the squares of the length of the chord intercepted by the line x+y=n $n$....

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 1

Here's a sketch of the situation:

We note that there are 2 lines having n as an integer and secant to the circle. Their equations are in the third quadrant.
We need to find the intersection points of the circle and each line, so we set up the following system:

$\begin{cases} x^2 + y^2 = 4 \\ x + y = 1 \end{cases}$

We don't need a system for the second line since we already know it passes through (0,2) and (2,0).
Once you get the coordinates, simply plug each pair into the distance formula, square each (or do a shortcut and just don't compute the square root over everything), and add it all together. That will be your answer.

Answer 2

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?

Answer 3

Referring to the diagram below:-

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.