Problem 70 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least 3 of these points is (A) 116 (B) 120 (C) 117 (D) None of these

Step-by-Step Solution

Answer

The correct answer is (C) 117.

Step 1: Understanding the Problem

We have 10 points in a plane, and no three of them are collinear. Among these, 4 points are concyclic, meaning they lie on the same circle. Our task is to find the number of different circles we can draw through at least 3 of these points.

Step 2: Counting All Possible Circles

First, count the ways to choose any 3 points out of the 10 to form a circle. Since no three points are collinear, a circle is formed by any three points.The number of ways to choose 3 points out of 10 is given by the combination formula: \[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

Step 3: Accounting for Concyclic Points

Next, account for the fact that 4 points are concyclic, meaning they can all be on the same circle. From these 4 points, any subset of 3 points will form the same circle. Therefore, we need to subtract the extra circles counted among these 4 points.

Step 4: Counting Ways for Concyclic Points

For the 4 concyclic points, we calculate the extra circles wrongly counted.The number of circles formed by any 3 points from these 4 is:\[ \binom{4}{3} = 4 \]However, these 4 combinations make only 1 unique circle, so we subtract 3 extra counts.

Step 5: Calculating the Total Number of Circles

Subtract the 3 overcounts from the total combinations.The total number of unique circles is then:\[ 120 - 3 = 117 \]

Step 6: Finalizing the Solution

Thus, the number of different circles that can be drawn through at least 3 of these points is 117.

Key Concepts

Concyclic Points

The term "concyclic points" refers to a set of points that all lie on the same circle. In simpler terms, if you can draw a single circle that passes through every point in a group, then those points are concyclic. This concept is significant in geometry because it helps us understand how points are distributed around a circle and helps in solving many geometric problems, such as those involving circle theorems.

In the context of our exercise, knowing that four of the ten points are concyclic means they can all simultaneously lie on a single circle. This characteristic must be considered while calculating total circles, as choosing any three points out of these four would result in the same circle being counted multiple times. This is why concyclic points play a pivotal role in accurately computing the number of unique circles possible.

Combination Formula

The combination formula is a fundamental tool in combinatorics for finding how many ways a given number of items can be selected from a larger group, disregarding the order of selection. The formula is typically expressed as:

• \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

where \( n \) is the total number of items to choose from, \( r \) is the number of items to select, and \( ! \) denotes factorial.

In the problem at hand, we use the combination formula to determine how many sets of 3 points can be selected from a pool of 10 points. Applying this formula, we calculate:

• \[ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

This computation reveals that there are 120 possible circles if no points were concyclic. However, due to the presence of concyclic points, further adjustment in this count is needed.

Circle Geometry

Circle geometry involves studying the properties and relations of points, lines, and angles associated with circles. One of its fundamental principles is that any three non-collinear points in a plane can define a unique circle. This property is the basis of our exercise in counting possible circles.

Typically, when dealing with circle geometry in combinatorics problems, we consider how different groupings of points can form circles and focus on issues like concyclic points and unique circle formation. For instance, in our exercise, because any set of three points forms a circle, we initially counted all combinations of three points to get 120 circles. However, because of the concyclic nature of some points, multiple combinations yield the same circle, requiring us to adjust for any overcounts. This highlights the importance of understanding these geometric relationships in practical problem solving.