This is how I went about doing it. I know that if there are 'n' number of straight lines they intersect each other in nC2 ("n choose 2") ways. Therefore, the points of intersection using 6 lines is 15.
I also know that a straight line and a circle intersect each other at atmost two different points. Hence the points of intersection of six lines and five circles is 6C1 x 5C1 x 2 = 60.
After this point I am a little confused. My answer was present in the options but it was the wrong one. I calculated it as being 15+60=75. But the answer key says it is 95. Their reason being that 2 circles intersect in atmost two different points and therefore the points of intersection of 5 circles is 5C2 x 2 = 20. The answer then works out to 15+60+20 = 95.
I did not understand the last part. Could someone please help me clear out my confusion? Thanks.
Best Answer
There are exactly $3$ types of intersection points:
Summing everything together, we obtain $15 + 60 + 20 = 95$.