There are 20 points of which 10 are collinear. Find the maximum number of circles that can be drawn.

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There are 20 points of which 10 are collinear. Find the maximum number of circles that can be drawn.

Mathematics
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\(\large \color{black}{\begin{align} & \normalsize \text{There are 20 points of which 10 are collinear.}\hspace{.33em}\\~\\ & \normalsize \text{Find the maximum number of circles that can be drawn.}\hspace{.33em}\\~\\ & a.)\ 900 \hspace{.33em}\\~\\ & b.)\ 800 \hspace{.33em}\\~\\ & c.)\ 700 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \end{align}}\)
is it none of these cuz max no circles can be drawn from any two points
  • phi
you left out some details. are the points the center, or do they lie on the circumference?

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Other answers:

that info is not given
  • phi
In that case, I have no idea.
btw 20C1*20C2+20C2*20C1=900 is given correct
  • phi
I guess they don't explain that ?
no
  • phi
maybe someone else will have some insight on this one.
  • phi
btw, 20C1*20C2+20C2*20C1 would work out to 20*190 + 190*20 that is not 900
btw book has many typos in the past
sry it is 10C1*10C2+10C2*10C1=900
  • phi
it sounds like they want the circle to go through 3 points
  • phi
a circle can be defined by 3 non-collinear points. If we pick 2 points from the line, then to get a circle we have to pick one of the points not on the line that must be how the get the 10C2 * 10C1 term
i want to make sure if the 2 points circle is not what they want
  • phi
you can define a unique circle if you give its center, and a point on the circumference or 3 points on the circumference only the 2nd interpretation makes sense with the answer. (there are an infinite number of circles that have the same two points on their circumference)
ok thnks
  • phi
I am only guessing on this. There really should be more info on what they are assuming. the other term is consistent with choosing 1 point on the line, and two points from the other set. but we could also choose all 3 points from the non-line set 10C3 and they leave that term out. So I don't really know what is going on.
u mean it should be \(900+10C3\)
  • phi
yes, unless you insist the circle must contain a point on the line.
ok i assume ur answer is correct then
With the crude information given, I would be tempted to give the following as answer: We have 20 points out of which any three should make a circle, N1=C(20,3)=1140 Out of these, the invalid ones are the choices which have all three falling on the collinear set of 10, so N2=C(10,3)=120 Hence the number of valid circles would be N1-N2=1140-120=\(1020\) This is also equal to include C(10,1)*C(10,2)+C(10,2)*C(10,1)+C(10,3)=450+450+120=1020 The last 120 are those not using the collinear points. That is also why @Phi was expecting more information/restrictions if 900 is the answer. I also believe either the question is incomplete, or the answer did not choose all possibilities.

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