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#### A plane contains 14 points, of which 4 are concyclic and none of which have three collinear points. Determine the number of different circles that can be drawn through at least three points of these points isOption: 1 361Option: 2 251Option: 3 441Option: 4 531

Given that,

There are 14 points in a plane, out of which 5 are concyclic and none of which have three collinear points.

The circles can be formed by joining 3 non-collinear points out of 14 points given by,

\mathrm{\begin{aligned} &{ }^{14} C_3=\frac{14 !}{3 ! 11 !}\\ &{ }^{14} C_3=\frac{14 \times 13 \times 12}{3 \times 2}\\ &{ }^{14} C_3=364 \end{aligned}}

The circles are also formed by joining 3 points out of 4 concyclic points is given by,

\mathrm{\begin{aligned} &{ }^4 C_3=\frac{4 !}{3 ! 1 !}\\ &{ }^4 C_3=4 \end{aligned}}

Also, we need to add 1 as there is a circle that would pass all 4 points.

Thus, the number of different circles that can be drawn through at least three points of these points is given by,

$364-4+1=361$

Therefore, the number of different circles formed is 361.