Question

There are $10$ points in a plane of which no $3$ points are collinear and $4$ points are concylic. No. of different circles that can be drawn through atleast $3$ points of these given points is

• A. $\left({}^8{C}_3{-}^6{C}_3\right)+1$
• B. $\left({}^{10}{C}_3{-}^4{C}_3\right)+1$
• C. $\left({}^6{C}_3{-}^4{C}_3\right)+1$
• D. $\left({}^4{C}_3{-}^2{C}_3\right)+1$

Answer 1

Answer: (B)

$\left({}^{10}{C}_3{-}^4{C}_3\right)+1$

We need at least $3$ points to form a circle:${}^{10}{C}_3$

Out of these $4$ points are already concyclic so we must remove ${}^4{C}_3-1$ cases.

${}^{10}{C}_3-({}^4{C}_3-1)$

Answer is:$\left({}^{10}{C}_3{-}^4{C}_3\right)+1$