Question

The line $x+y=n$, $n\in N$, make intercepts with $x^2+y^2=16$. Then the sum of squares of all possible intercepts is

• A. $\frac{105}{4}$
• B. $105$
• C. $210$
• D. $\frac{105}{2}$

Answer 1

The correct option is C

$210$

Step 1: Find the possible value of $n$.

A line $x+y=n$, $n\in N$ and a circle $x^2+y^2=16$ are given.

It is clear that the centre of the given circle is $(0,0)$ and the radius is $4$.

We know that, if $Ax+By+c=0$ is a line and $(x_1,y_1)$ is a point then the perpendicular distance $D$ between the line and the point is given by: $D=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}$.

Since the perpendicular distance between the line and the centre of the circle cannot be more than the radius of the circle to make an intercept at two different points.

So,

$$\begin{gathered} \frac{\left|0+0-n\right|}{\sqrt{1^2+1^1}}<4 \\ \Rightarrow \frac{n}{\sqrt{2}}<4 \\ \Rightarrow n<4\sqrt{2} \end{gathered}$$

So, the possible values of $n$ such that $n\in N$ are $\left\{1,2,3,4,5\right\}$.

Step 2: Find the sum of squares of all possible intercepts.

we know that the length of the intercept of a circle can be given by $2\sqrt{r^2-p^2}$ where $r$ is the radius of the circle and $p$ is the perpendicular distance between the center of the circle and the line.

So, the length $l$ of the intercept is $2\sqrt{16-\frac{n^2}{2}}$.

So,

$$\begin{gathered} \sum _{n=1}^5{l}^2=4\left[\left(16-\frac{1}{2}\right)+\left(16-\frac{4}{2}\right)+\left(16-\frac{9}{2}\right)+\left(16-\frac{16}{2}\right)+\left(16-\frac{25}{2}\right)\right] \\ \Rightarrow \sum _{n=1}^5{l}^2=4\left[\frac{31}{2}+14+\frac{23}{2}+8+\frac{7}{2}\right] \\ \Rightarrow \sum _{n=1}^5{l}^2=4\left[52.5\right] \\ \Rightarrow \sum _{n=1}^5{l}^2=210 \end{gathered}$$

Therefore, the sum of squares of all possible intercepts is $210$.

Hence, the correct answer is option C.