Question

There are 8 points on a plane, out of which 4 points lies on the circumference of the same circle and rest 4 points do not lie on a single circumference of a circle and also they are non-collinear. Maximum how many circles can be drawn such that each contains at least three of the given points?

• A. 53
• B. 32
• C. 35
• D. 56

Answer 1

The correct option is A 53

Maximum probable number of circles = ${}^8{C}_3$
(Since a circle can pass through any three non-collinear points)
But since 4 points lie on the same circle so it reduces the formation of some circles.
$\therefore$ Actual number of circles ${=}^8{C}_3{-}^4{C}_3+1$
= 56 - 4 + 1 = 53
Note: Since when we subtract ${}^4{C}_3$, we actually reduce one possible circle, so we have to add 1 back.