Question

If there are $n$ circles in a plane which gives $56$ points of intersection, then the minimum value of $n$ is

• A. $7$
• B. $8$
• C. $10$
• D. $12$

Answer 1

The correct option is B $8$

Two circles intersect at atmost two points.
So, the maximum number of points of intersection of $n$ circles in a plane is $2\times {}^n{C}_2$
$\therefore 56\le 2\times {}^n{C}_2$
or, $2\times {}^n{C}_2\ge 56$
$\Rightarrow 2\times \frac{n(n-1)}{2}\ge 56$
$\Rightarrow n^2-n-56\ge 0$
$\Rightarrow (n-8)(n+7)\ge 0$
$\Rightarrow n_{min}=8$