Question

In $n(n\ge 3)$ circles the centres of no three circles are collinear. If 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

Answer: (B)

$0$

Radical axis is defined for every pair of circles.

So number of radical axes $={}^n{C}_2$

Radical center is defined for every three circles.

So the number of radical centres is ${}^n{C}_3$

Since number of radical axes is equal to number of radical centers, we have ${}^n{C}_2={}^n{C}_3$

$\Rightarrow n-2=3\Rightarrow n=5$

Therefore, $n^2-4n-5=0$

So the right answer is option B.