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I came across this question in my class:

There are 11 different points in the plane with no 3 points are on the same line.
a) How many circles do these points define? (Points define a circle if there is a unique circle through those points.)
b) How many circles would they define, if 4 points were on the same line?

I think, that we just need 2 points, to define a circle (one for the centre and 1 for the radius). In that case a) would be $11\times10=110$ different circles, however that seems to be incorrect.

How would you solve it?

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    $\begingroup$ Nope, try with 3 points. $\endgroup$ – The Dead Legend May 10 '17 at 5:57
  • $\begingroup$ Wait, are there 11 or 12 points? $\endgroup$ – Parcly Taxel May 10 '17 at 5:58
  • $\begingroup$ @TheDeadLegend why 3 points? $\endgroup$ – Marinaro May 10 '17 at 6:01
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    $\begingroup$ Are all $11$ points allowed to be on the same circle? The answer would be different vs. the case where no $4$ points are concyclic. $\endgroup$ – dxiv May 10 '17 at 6:01
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    $\begingroup$ @WannaBeGnome If the question doesn't specify it, you can't assume it just because it's easier to solve it that way. In that case, the answer would not be a single number, but a set (or range) of possible numbers. $\endgroup$ – dxiv May 10 '17 at 6:09
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It doesn't take two points to define a circle, because either of those two points could be the centre. Instead, the solution for part (a) is simply $\binom{11}3=165$ because three non-collinear points uniquely define a circle (assuming that multiplicity is counted, or that no four points are concyclic; if such a quartet existed the number of distinct circles would be lower).

For part (b), $\binom43=4$ circles become degenerate and need to be subtracted, so the answer is 161 circles in this case (again, with one of the assumptions above).

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  • $\begingroup$ I agree with $\binom{11}{3}$ but it still seems like you can create two circles per pair as described by the OP which would add $2\cdot \binom{11}{2}$ to this total. $\endgroup$ – law-of-fives May 10 '17 at 6:03
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    $\begingroup$ A family of circle is described about 2 points. To get a particular circle, we need 3 points @law-of-fives $\endgroup$ – The Dead Legend May 10 '17 at 6:04
  • $\begingroup$ @TheDeadLegend yes, of course, I understand. Sometimes I'm slow! $\endgroup$ – law-of-fives May 10 '17 at 6:05
  • $\begingroup$ Thanks. I didn´t realise that 2 points aren't enough to define a circle. Thanks for an explanation. $\endgroup$ – Marinaro May 10 '17 at 6:09
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    $\begingroup$ @WannaBeGnome, for clarity, 2 points is enough to define a circle if you add some extra rules like which to take as a centre or how to define the radius. However, your own questions states "Points define a circle if there is a unique circle through those points", so then you need 3 points to uniquely define a circle. $\endgroup$ – Wolfie May 10 '17 at 8:48
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Hint: $3$ points define a circle

$^{11}C_3$ for case 1 because no three in same line

$^{11}C_3-^4C_3$ for case 2, to exclude cases taking three points from those 4 collinear points

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  • $\begingroup$ (taking them to be 11 points in total) $\endgroup$ – The Dead Legend May 10 '17 at 6:00

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