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If 6 circles are drawn in a plane, what is the max number of points of intersection of these 6 circles?
each circle can intersect another circle in 2 points, hence the number of points of intersection for 6 circles
= 2*6C2= 2* 6*5/2= 30

hence D is the answer
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a circle intersects another circle with 2 points. if we add another circle, first circle again intersects with 2 points at max.
from this
circle (1) intersects 5 other circles with each 2 points: 5*2=10 points
circle (2) intersects 4 other circle with each 2 points: 4*2=8 points /because we counted before the points that created intersection between circle (1) and (2), so we do not have to count again/
circle (3) intersects 3 other circle with each 2 points: 3*2=6 points
this continues until circle (5) and (6)
so the total intersection points would be 5*2+4*2+3*2+2*2+1*2=30
D is right
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By assuming, one circle has one line. In case of N lines on the plain, it's always possible to add one more that will intersect all of them, and thus create N new intersection points. All that it is required to occur is to ensure that the added line is not parallel to any of the existing lines, and that it doesn’t intersect any of the lines at an existing intersection.
Just setting a new line at random will almost always comply with these requirements.

Based on that, we can calculate the maximum number of intersection for N lines:

0+1+2+3+…+(N−1)=N(N−1)/2=N^2−N/2

And specifically, if N=6, then the the maximum number of intersections is:

N(N−1)/2=6x5/2=15

or

It also can be solved using combination formula.
Two circles can intersect in at most two points (excluding the case where two circles coincide with each other). So by looking at all possible pairs of circles, the maximum number of intersection points is given by 6C2 = 6x5/2=15.

Ans. A.
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If 6 circles are drawn in a plane, what is the max number of points of intersection of these 6 circles?
A. 15
B. 20
C. 25
D. 30
E. 38

Answer - D

For every 2 circles, there are maximum 2 points of intersections.

Hence Selecting any 2 circles from 6 circles => \(6_C_2\) and for every pair we have 2 intersections => \(6_C_2\) * 2

Therefore total = 6*5 = 30
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one circle intersects with other at max 2 times
and one circle can interests with 5 others circles
hence
total intersections
= \(\frac{6*5*2}{2}\)
=30
D
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If 6 circles are drawn in a plane, what is the max number of points of intersection of these 6 circles?

suppose there are 2 circles
maximum number of intersection between 2 circles = 2

now, there are 6 circles
number of ways of selection of 2 circles from 6 = 6c2 = \(\frac{6!}{4!*2!}\) = 15

so, there are total 15 pairs ---maximum number of intersection of each pair of circles = 2
since there are 15 pairs --maximum number of intersection = 15*2 = 30

max number of points of intersection of 6 circles = 30

D is the answer
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This is a tricky one.
Attachments

New Doc 2020-02-16 08.09.13_6.jpg
New Doc 2020-02-16 08.09.13_6.jpg [ 205.91 KiB | Viewed 5697 times ]

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Solution:

Question 7: If 6 circles are drawn in a plane, what is the max number of points of intersection of these 6 circles?

A. 15
B. 20
C. 25
D. 30
E. 38

We know that any two circles can intersect at maximum 2 points. In this question, we need to find out the maximum number of points of intersection of six circles. Since any 2 circles can intersect at maximum two points, any 2 circles out of 6 circles will also intersect at 2 points.

Thus, the maximum number of intersection points are \(6C_{2} * 2\) = 30, which is option D.

Therefore, option D is the correct answer.
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Not sure, but the answer could be C or D, depending on how I drew the circles.
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