Question

The sum of the square of the chord intercepted by the line $x+y=n,n\epsilon N$ on the circle $x^2+y^2=4$ is

• A. 11
• B. 22
• C. 33
• D. none of these

Answer 1

The correct option is B 22

The given equation of the circle is,

$x^2+y^2=4$

$x^2+y^2={\left(2\right)}^2$

Thus, coordinates of center of circle are $O(0,0)$ and radius is $r=2$

1) Now, the equation of the line is given as,

$x+y=n$

Put n=1 in above equation, we get first equation of straight line.

$\therefore x+y=1$

Put $y=0$ in above equation, we get

$x+0=1$

$\therefore x=1$

Similarly, put $x=0$ in above equation ,we get

$0+y=1$

$\therefore y=1$

Thus, this line passes through points $(1,0)$ and $(0,1)$

2) To find point of intersection of line with circle-

Equation of line is given by, $x+y=1$

$\therefore y=1-x$

Put this value in equation of circle, we get,

$x^2+{(1-x)}^2=4$

$\therefore x^2+1-2x+x^2=4$

$\therefore 2x^2-2x-3=0$

$\therefore x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4\times 2\times (-3)}}{2\times 2}$

$\therefore x=\frac{2\pm \sqrt{4+24}}{4}$

$\therefore x=\frac{2\pm \sqrt{28}}{4}$

$\therefore x=\frac{2\pm \sqrt{4\times 7}}{4}$

$\therefore x=\frac{2\pm 2\sqrt{7}}{4}$

$\therefore x=\frac{1\pm \sqrt{7}}{2}$

$\therefore x=\frac{1+\sqrt{7}}{2}$ and $x=\frac{1-\sqrt{7}}{2}$

When $x=\frac{1+\sqrt{7}}{2}$, $y=1-\frac{1+\sqrt{7}}{2}$

$\therefore y=\frac{2-(1+\sqrt{7})}{2}$

$\therefore y=\frac{(1-\sqrt{7})}{2}$

When $x=\frac{1-\sqrt{7}}{2}$, $y=1-\frac{1-\sqrt{7}}{2}$

$\therefore y=\frac{2-(1-\sqrt{7})}{2}$

$\therefore y=\frac{(1+\sqrt{7})}{2}$

Thus, points of intersection of line $L_1$ with circle are $A(\frac{1+\sqrt{7}}{2},\frac{1-\sqrt{7}}{2})$ and $B(\frac{1-\sqrt{7}}{2},\frac{1+\sqrt{7}}{2})$

Thus, length of chord AB by distance formula is,

$AB=\sqrt{{(\frac{1-\sqrt{7}}{2}-\frac{1+\sqrt{7}}{2})}^2+{(\frac{1+\sqrt{7}}{2}-\frac{1-\sqrt{7}}{2})}^2}$

$AB=\sqrt{{\left(\frac{1-\sqrt{7}-1-\sqrt{7}}{2}\right)}^2+{\left(\frac{1+\sqrt{7}-1+\sqrt{7}}{2}\right)}^2}$

$AB=\sqrt{{\left(\frac{-2\sqrt{7}}{2}\right)}^2+{\left(\frac{2\sqrt{7}}{2}\right)}^2}$

$AB=\sqrt{{(-\sqrt{7})}^2+{\left(\sqrt{7}\right)}^2}$

$\therefore AB=\sqrt{7+7}$

$\therefore AB=L_1=\sqrt{14}$

3) put n=2 in equation of line.

$\therefore x+y=2$

$\therefore y=2-x$

Put y=0 in above equation. we get

$x+0=2$

$\therefore x=2$

Put x = 0 in above equation, we get,

$0+y=2$

$\therefore y=2$

The line passes through points $D(2,0)$ and $E(0,2)$ and intercepts circle with chord DE.

As shown in figure, by distance formula,

$DE=\sqrt{{(2-0)}^2+{(0-2)}^2}$

$\therefore DE=\sqrt{{\left(2\right)}^2+{(-2)}^2}$

$\therefore DE=\sqrt{4+4}$

$\therefore DE=L_2=\sqrt{8}$

For $n\ge 3$, lines will not intercept with circle. Thus, only two lines are considered as chords.

Now, Sum of square of chords is ${L_1}^2+{L_2}^2$

$\therefore {L_1}^2+{L_2}^2={\left(\sqrt{14}\right)}^2+{\left(\sqrt{8}\right)}^2$

$\therefore {L_1}^2+{L_2}^2=14+8$

$\therefore {L_1}^2+{L_2}^2=22$

Thus, answer is option (B)