mathmath333
  • mathmath333
There are 20 points of which 10 are collinear. Find the maximum number of circles that can be drawn.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{There are 20 points of which 10 are collinear.}\hspace{.33em}\\~\\ & \normalsize \text{Find the maximum number of circles that can be drawn.}\hspace{.33em}\\~\\ & a.)\ 900 \hspace{.33em}\\~\\ & b.)\ 800 \hspace{.33em}\\~\\ & c.)\ 700 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \end{align}}\)
mathmath333
  • mathmath333
is it none of these cuz max no circles can be drawn from any two points
phi
  • phi
you left out some details. are the points the center, or do they lie on the circumference?

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More answers

mathmath333
  • mathmath333
that info is not given
phi
  • phi
In that case, I have no idea.
mathmath333
  • mathmath333
btw 20C1*20C2+20C2*20C1=900 is given correct
phi
  • phi
I guess they don't explain that ?
mathmath333
  • mathmath333
no
phi
  • phi
maybe someone else will have some insight on this one.
phi
  • phi
btw, 20C1*20C2+20C2*20C1 would work out to 20*190 + 190*20 that is not 900
mathmath333
  • mathmath333
btw book has many typos in the past
mathmath333
  • mathmath333
sry it is 10C1*10C2+10C2*10C1=900
phi
  • phi
it sounds like they want the circle to go through 3 points
phi
  • 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
mathmath333
  • mathmath333
i want to make sure if the 2 points circle is not what they want
phi
  • 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)
mathmath333
  • mathmath333
ok thnks
phi
  • 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.
mathmath333
  • mathmath333
u mean it should be \(900+10C3\)
phi
  • phi
yes, unless you insist the circle must contain a point on the line.
mathmath333
  • mathmath333
ok i assume ur answer is correct then
mathmate
  • mathmate
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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