Question

The maximum number of points of intersection of five lines and four circles in a plane is

• A. $60$
• B. $62$
• C. $70$
• D. $72$

Answer 1

The correct option is B $62$

Maximun number of points of intersection of
$\left(i\right)$ $n$ circles in a plane is $2{\times }^n{C}_2$.
$\left(ii\right)$$m$ straight lines in a plane is ${}^m{C}_2$
$\left(iii\right)$ $n$ circles and $m$ straight lines in a plane is $2{\times }^n{C}_1{\times }^m{C}_1$

$\therefore$ Required number of points of intersection are as follows:
Number of ways of selection Points of intersection

Two straight lines : ${}^5{C}_2$	${}^5{C}_2\times 1=10$
Two circles : ${}^4{C}_2$	${}^4{C}_2\times 2=12$
One line and one circle ${}^5{C}_1\times {}^4{C}_1$	${}^5{C}_1\times {}^4{C}_1\times 2=40$
Total	$=62$