Question:
There are 4, 5 and 6 points on the three parallel and co-planar lines. If no set of four or more points is cyclic and no set of three points lying on distinct lines is collinear, then the number of circles passing through exactly three points is :
There are 4, 5 and 6 points on the three parallel and co-planar lines. If no set of four or more points is cyclic and no set of three points lying on distinct lines is collinear, then the number of circles passing through exactly three points is :
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Any three non-collinear points determine exactly one circle, so count all triples and remove the collinear ones.
Updated On: Nov 12, 2026
- 455
- 421
- 435
- 441
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The Correct Option is B
Solution and Explanation
Total points = 4 + 5 + 6 = 15, so the number of triples is \( \binom{15}{3}=455 \).
Three points on the same line are collinear and give no circle. Number of such triples: \( \binom43+\binom53+\binom63=4+10+20=34 \).
Triples with points on distinct lines are non-collinear (given), and triples with two points on one line and one on another are also non-collinear, so each gives exactly one circle. No four points are concyclic, so no circle is counted twice.
Number of circles \( =455-34=421 \)
Answer: (2) 421
Three points on the same line are collinear and give no circle. Number of such triples: \( \binom43+\binom53+\binom63=4+10+20=34 \).
Triples with points on distinct lines are non-collinear (given), and triples with two points on one line and one on another are also non-collinear, so each gives exactly one circle. No four points are concyclic, so no circle is counted twice.
Number of circles \( =455-34=421 \)
Answer: (2) 421
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