Question

There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least 3 points of these points, is

• A. $116$
• B. $117$
• C. $120$
• D. None of these

Answer 1

The correct option is B $117$

Given data:

No of points in a plane$=10$

No of non collinear points$=3$

No of concyclic points$=3$

To find:

No of different circles drawn through at least three points

Solution:

No of circles $={}^{10}{C}_3-{}^4{C}_3+1$

$=\frac{10\times 8\times 9}{1\times 2\times 3}$

$=120-4+1$

$=117$