Question

The sum of the squares of the length of the chords intercepted on the circle, ${{x}^{2}}+{{y}^{2}}=16$ , by the lines $x+y=n,\text{ }n\in N$ where $N$ is the set of all natural number , is:
(a) 320
(b) 160
(c) 105
(d) 210

Answer 1

Hint: $Length\text{ }of\text{ }chord=2\sqrt{16-\dfrac{{{n}^{2}}}{2}}$ .

${{\left( Length\text{ }of\text{ }chord \right)}^{2}}=2\left( 32-{{n}^{2}} \right)$ . Possible value of $n=1,2,3,4,5$ . Put the value of n and then add the squares of the length of chords.

Complete step by step answer:

First of all, let us draw a figure to understand how to find the length of a chord.

This is the figure of circle ${{x}^{2}}+{{y}^{2}}=16$ and lines $x+y=n$ intercepts the circle; we have equation of circle, ${{x}^{2}}+{{y}^{2}}=16$ i.e., ${{x}^{2}}+{{y}^{2}}={{\left( 4 \right)}^{2}}$ , where $radius\text{ }OA=radius\text{ }OB=4$ .

The centre of the circle is $O\left( 0,0 \right)$ .

We will find the length of $AB$ .

Also we know that the perpendicular drawn from the centre of a circle to the chord of that circle bisects the chord.

So we have

$AB=2\left( AN \right)$

Now let us understand how to find the perpendicular distance from the centre of a circle to the chord of that circle.

If $d$ is the perpendicular distance from the centre $P\left( {{x}_{1}},{{y}_{1}} \right)$ of the circle to the chord $ax+by+c=0$ , then

$d=\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$

Here, we have , $centre\text{ }of\text{ }circle=O\left( 0,0 \right)$ and equation of chord is $x+y=n$ i.e., $x+y-n=0$ . $ON$ is the perpendicular distances from the centre to the chord of the circle.

$\therefore ON=\left| \dfrac{1.0+1.0+(-n)}{\sqrt{{{1}^{2}}+{{1}^{2}}}} \right|$

$\Rightarrow ON=\left| \dfrac{0+0-n}{\sqrt{1+1}} \right|$

$\Rightarrow ON=\left| \dfrac{-n}{\sqrt{2}} \right|$

As $n$ is a natural number therefore, $n>0$ .

Thus, $\left| -n \right|=n$ .

$\therefore ON=\dfrac{n}{\sqrt{2}}$

In the figure, we can see that $ON$ is perpendicular and $OA$ is hypotenuse.

So, $ON < OA$ , $OA$ being radius

i.e., $\dfrac{n}{\sqrt{2}}<4$

i.e., $n<4\sqrt{2}$

Now we know that $\sqrt{2}=1.41$

$\therefore n<4\times 1.41$

i.e., $n<5.64$

Thus, the possible values of $n$ are $1,2,3,4,5$ .

Now take a look at the figure.

$\Delta OAN$ is a right-angled triangle.

Therefore by Pythagoras theorem we have

$\text{ }{{\left( OA \right)}^{2}}={{\left( AN \right)}^{2}}+{{\left( ON \right)}^{2}}$

$\Rightarrow {{\left( 4 \right)}^{2}}={{\left( AN \right)}^{2}}+{{\left( \dfrac{n}{\sqrt{2}} \right)}^{2}}$

$\Rightarrow 16={{\left( AN \right)}^{2}}+\dfrac{n}{2}$

$\Rightarrow {{\left( AN \right)}^{2}}=16-\dfrac{{{n}^{2}}}{2}$

$\Rightarrow AN=\sqrt{16-\dfrac{{{n}^{2}}}{2}}$

We know that

$AB=2\left( AN \right)$

$\Rightarrow AB=2\sqrt{16-\dfrac{{{n}^{2}}}{2}}$

Here $AB$ is the length of chord. For finding the square of length of the chord, we will take square on both sides. Then we have

$A{{B}^{2}}={{\left( 2\sqrt{16-\dfrac{{{n}^{2}}}{2}} \right)}^{2}}$

$\Rightarrow A{{B}^{2}}=4\left( 16-\dfrac{{{n}^{2}}}{2} \right)$

$\Rightarrow A{{B}^{2}}=4\dfrac{\left( 32-{{n}^{2}} \right)}{2}$

$\Rightarrow A{{B}^{2}}=2\left( 32-{{n}^{2}} \right)$

Now we will put the values of n and find the sum of squares of the length of chords. As we have $n=1,2,3,4,5$ therefore the sum of squares of the length of the chord

$=2\left( 32-{{1}^{2}} \right)+2\left( 32-{{2}^{2}} \right)+2\left( 32-{{3}^{2}} \right)+2\left( 32-{{4}^{2}} \right)+2\left( 32-{{5}^{2}} \right)$

$=2\left( 32-1 \right)+2\left( 32-4 \right)+2\left( 32-9 \right)+2\left( 32-16 \right)+2\left( 32-25 \right)$

$=2\times 31+2\times 28+2\times 23+2\times 16+2\times 7$

$=62+56+46+32+14$

$=210$

Thus the sum of the squares of the length of the chords intercepted on the circle ${{x}^{2}}+{{y}^{2}}=16$ , by the lines $x+y=n,\text{ }n\in N$ , where $N$ is the set of natural numbers , is $210$ .

So, the correct answer is “Option D”.

Note: The student must remember the formulas of finding the particular distance from the centre of the circle to the chord of that circle. One can go wrong in finding the value of $n$ if he/she is not able to identify hypotenuse and perpendicular in a triangle.