Question

Consider the set $A_n$ of points $(x,y)$ such that $0\le x\le n,0\le y\le n$ where $n,x,y$ are integers. Let $S_n$ be the set of all lines passing through at least two distinct points from $A_n$. Suppose we choose a line $l$ at random from $S_n$. Let $P_n$ be the probability that $l$ is tangent to the circle $x^2+y^2=n^2(1+{(1-\frac{1}{\sqrt{n}})}^2)$. Then the limit $\underset{n\to \infty }{lim}{P}_n$ is

• A. 0.0
• B. 1
• C. $\frac{1}{\pi }$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A 0.0

Assuming two points $(x_1,y_1)$ and $(x_2,y_2)$
The line joining the two points will be equal,
$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\Rightarrow (x_2-x_1)y-(x_2-x_1)y_1=(y_2-y_1)x-(y_2-y_1)x_1\Rightarrow (y_1-y_2)x+(x_2-x_1)y-(x_2-x_1)y_1+(y_2-y_1)x_1=0\Rightarrow ax+by+c=0$
Square of the distance of the line from the centre of the circle $(0,0)$ will be
$d^2=\frac{c^2}{a^2+b^2}$
This is a rational number.
There are two cases possible

Case 1: When the value of $n$ is not a perfect square then,
$r^2=n^2(1+{(1-\frac{1}{\sqrt{n}})}^2)$
So this will be irrarational,
Therefore the distance cannot be equal to radius, so the given line will never be tangent to the circle.

Case 2: $n$ is a perfect square.
$r^2=n^2(1+{(1-\frac{1}{\sqrt{n}})}^2)$
The number of tangents possible is very few compared to the total number of lines possible.
So either no tangent possible or very few tangent possible.

Hence, $\underset{n\to \infty }{lim}{P}_n=0$