Given circle and family of lines

The circle is $x^2+y^2=16$ so its center is $(0,0)$ and radius is $r=4.$

The lines are

Condition for a line to cut a chord in the circle

A line intersects the circle in a real chord if its perpendicular distance from the center is at most the radius.

For the line $x+y-n=0,$ the distance from $(0,0)$ is $d=\frac{|n|}{\sqrt{1^2+1^2}}=\frac{n}{\sqrt{2}}$ since $n\in\mathbb{N}$.

For intersection, $\frac{n}{\sqrt{2}}\le 4$ $n\le 4\sqrt{2}.$

Since $n$ is a natural number, $n=1,2,3,4,5.$

Length of chord cut by the line $x+y=n$

For a circle of radius $r$, if the distance of the chord from the center is $d$, then chord length is $L=2\sqrt{r^2-d^2}.$

Hence, $L_n=2\sqrt{16-\frac{n^2}{2}}.$

So the square of the chord length is $L_n^2=4\left(16-\frac{n^2}{2}\right)=64-2n^2.$

Required sum

We need $\sum_{n=1}^{5} L_n^2=\sum_{n=1}^{5}(64-2n^2).$

Now, $\sum_{n=1}^{5}64=5\cdot 64=320$ and $\sum_{n=1}^{5}n^2=1^2+2^2+3^2+4^2+5^2=1+4+9+16+25=55.$

Therefore, $\sum_{n=1}^{5} L_n^2 = 320-2(55)=320-110=210.$

Option check

• A: $210$ ✅
• B: $160$
• C: $320$
• D: $105$

Hence the correct option is A.