Question

In $n(n\ge 3)$ circles the centres of no three circles are collinear. lf the number of the radical axes of the circles is equal to the number of the radical centres of the circles then $n^2-4n-5=$

• A. $5$
• B. $0$
• C. $50$
• D. $7$

Answer 1

The correct option is B $0$

Number of radical axis ${=}^n{C}_2$ and number of radical centre ${=}^n{C}_3$
Given both are same, ${}^n{C}_2{=}^n{C}_3\Rightarrow n-2=3\Rightarrow n=5$
Hence $n^2-4n-5=25-20-5=0$