Find the maximum number of points of intersection of 6 circles.

Question

I know that the answer is 30, but how we are getting to it, that I do not know. One more thing that I know that the maximum number of intersection of 2 circles is 2.

Answer 1

If every pair of circles can intersect in at most $2$ points, then the max number of intersections among the $6$ circles is twice the number of pairs of circles we can form: $$2\binom{6}{2}=30$$ Edit: following the comment of @David K I will show that we can actually obtain this number. Consider the $6$ circles of radius $10$ centered at $(a,0$) for $a=0,1,2,3,4,5$. Then one can easily check that for all $a,b\in\{0,1,2,3,4,5\}$, the two points $$\left(\frac{a+b}{2},\pm\sqrt{100-\frac{(a-b)^{2}}{4}}\right)$$ lie on both circles $(x-a)^{2}+y^{2}=100$ and $(x-b)^{2}+y^{2}=100$. I'll mention that to find these points, I subtracted the equations of both circles to obtain $$-2bx+b^{2}-(-2ax+a^{2})=0$$ which reduces to $(b-a)(b+a-2x)=0$ and since $a\neq b$ for distinct circles, we must have $x=(a+b)/2$. It is also clear by looking at the $x$-coordinates of these intersection points that they are all distinct, so that the max of $30$ is indeed obtained.

Answer 2

Let’s place the first circle.
Now the second circle, which makes 2 intersections.
The third circle gives 4 more intersections.(2 intersections to each circle.)
That will give us 2+4+6+8+10=30 at the end.
I drew a picture with 30 intersections for you.