question_answer

The line \[x+y=n,n\in N\] makes intercepts with \[{{x}^{2}}+{{y}^{2}}=16.\] Then the sum of square of all possible intercepts.

A) \[\frac{105}{4}\]

B) \[105\]

C) \[210\]  

D) \[\frac{105}{2}\]

Correct Answer: D

Solution :

\[p=\frac{n}{\sqrt{2}}\]

to make the intercept,

\[\frac{n}{\sqrt{2}}<4\]\[\Rightarrow \]     \[n<4\sqrt{2}\]

Length of intercepts\[=\sqrt{{{r}^{2}}-{{p}^{2}}}\]\[=\sqrt{16-{{n}^{2}}/2}\]

Square of intercept\[=16-\frac{{{n}^{2}}}{2},n\in N\]

Sum of squares of intercept

\[\left( 16-\frac{1}{2} \right)+\left( 16-\frac{4}{2} \right)+\left( 16-\frac{9}{2} \right)+\left( 16-\frac{16}{2} \right)+\left( 16-\frac{25}{2} \right)\]\[=80-\frac{1}{2}(55)\]\[=\frac{105}{2}.\]