Question

There are $10$ points in a plane of which no $3$ points are collinear and $4$ points are concyclic. Number of different circles that can be drawn through at least $3$ points is

• A. $100$
• B. $116$
• C. $117$
• D. $120$

Answer 1

The correct option is C $117$

Since there are $10$ points and no $3$ points are collinear and $4$ points are concyclic.

Since, there will be only one circle passing through this $4$ concyclic points, hence we will subtract these circles because they are repetitive. And now we will add $1$ as there is circle which would pass all $4$ points.
The number of required circles
$=({}^{10}{C}_3-{}^4{C}_3)+1$
$=117$