There are 12 points in a plane of which no three points are collinear and 5 points are concyclic. The number of different circles that can be drawn through atleast 3 points of these points is

Out of 12 points 5 points are concyclic.

Total number of circles that can be drawn using 12 points $=12{\mathrm{C}}_3$

Since 5 points are concyclic $5{\mathrm{C}}_3$ different circles are not possible but we can draw one circle.

 $\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{circles} & =& 12{\mathrm{C}}_3-5{\mathrm{C}}_3+1\\ & =& 220-10+1\\ & =& 211 \end{array}$