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
$\left(1\right)n$ circles in a plane is $2\times {}^n{C}_2$
$\left(2\right)m$ straight lines in a plane is ${}^m{C}_2$
$\left(3\right)n$ 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:
$\begin{array}{cc}\text{Number of ways of}& \text{Points of}\\ \text{selection}& \text{intersection}\\ \text{Two straight lines :}{}^5{C}_2& {}^5{C}_2\times 1=10\\ \text{Two circles :}{}^4{C}_2& {}^4{C}_2\times 2=12\\ \text{One line and one circle :}& {}^5{C}_1\times {}^4{C}_1\times 2\\ {}^5{C}_1\times {}^4{C}_1& =40\\ \text{Total}& =62\end{array}$