<p>There are 10 points in a plane of which no three points are collinear and four points are concyclic. The number of different circles that can be drawn through at least three points of these points is</p>
Step-by-Step Solution
Key Concept: To count circles through at least 3 points, use total combinations of 3 points minus those that are collinear, then subtract the concyclic 4-point group overcounting since they form only 1 circle, not C(4,3) circles.
<p><strong>Step 1:</strong> Find total combinations of 3 points from 10 points: C(10,3) = 120</p><p><strong>Step 2:</strong> Since no three points are collinear, all these combinations give valid circles (no collinear subtractions needed).</p><p><strong>Step 3:</strong> The 4 concyclic points are special. They can form C(4,3) = 4 different 3-point combinations, but all 4 combinations lie on the same single circle.</p><p><strong>Step 4:</strong> Subtract the overcounting: We counted 4 circles but there's only 1 actual circle. The overcount is 4 - 1 = 3.</p><p><strong>Step 5:</strong> Number of distinct circles = 120 - 3 = <strong>117</strong></p><p>∴ Answer: C</p>
Correct Answer: C