Question

The sum of the squares of the lengths of the chords intercepted on \(x^2+y^2=16\) by the lines \(x+y=n, n \in \mathbb{N}\) is:

• A. 320
• B. 210
• C. 180
• D. 120

Hint:

Use the summation formula \(\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}\) to quickly find the sum of squares for larger \(n\).

Answer 1

The Correct Option is B

Step 1: Picture the setup.
We have the circle $x^2+y^2=16$, whose centre is the origin and whose radius is $R=4$. A family of lines $x+y=n$ (with $n$ a natural number) cuts chords across it. We want the sum of the squares of all such chord lengths.
Step 2: Recall the half-chord idea.
Drop a perpendicular from the centre to a chord. It bisects the chord, so if the perpendicular distance is $d$, the half-chord is $\sqrt{R^2-d^2}$ and the full chord is $L=2\sqrt{R^2-d^2}$.
Step 3: Find the distance from the centre to a line.
For the line $x+y-n=0$, the distance from $(0,0)$ is \[ d=\frac{|{-n}|}{\sqrt{1^2+1^2}}=\frac{n}{\sqrt2}. \] So $d^2=\dfrac{n^2}{2}$.
Step 4: Write the square of the chord length.
\[ L^2=4\left(R^2-d^2\right)=4\left(16-\frac{n^2}{2}\right)=64-2n^2. \] This neat formula lets us just plug in each allowed $n$.
Step 5: Decide which values of $n$ are allowed.
The line only cuts a real chord when $d<R$, that is $\dfrac{n}{\sqrt2}<4$, giving $n<4\sqrt2\approx5.66$. Since $n$ is a natural number, $n=1,2,3,4,5$.
Step 6: Add them all up.
\[ \sum_{n=1}^{5}\left(64-2n^2\right)=5(64)-2\,(1+4+9+16+25). \] Here $1+4+9+16+25=55$, so the sum is $320-2(55)=320-110=210$.
So the total of the squared chord lengths is $210$.
\[ \boxed{210} \]