Problem 68 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If three distinct points \(A, B, C\) are given in the 2 -dimensional coordinate plane such that the ratio of the distance of each one of them from the point \((1,0)\) to the distance from \((-1,0)\) is equal to \(\frac{1}{2}\), then the circumcentre of the triangle \(A B C\) is at the point (a) \(\left(\frac{5}{3}, 0\right)\) (b) \((0,0)\) (c) \(\left(\frac{1}{3}, 0\right)\) (d) \((3,0)\)

Step-by-Step Solution

Answer

The circumcentre of triangle \( ABC \) is \( \left(\frac{5}{3}, 0\right) \).

Step 1: Understand the Given Condition

We know each point, say point P, satisfies the condition that the ratio of its distance from (1,0) to its distance from (-1,0) is \( \frac{1}{2} \). This can be expressed as \( \frac{\text{distance}(P, (1,0))}{\text{distance}(P, (-1,0))} = \frac{1}{2} \).

Step 2: Express Distances Using the Distance Formula

For a point \( P(x,y) \), the distance from \((1,0)\) is \( \sqrt{(x-1)^2 + y^2} \) and the distance from \((-1,0)\) is \( \sqrt{(x+1)^2 + y^2} \).

Step 3: Set Up the Ratio Equation

Substitute the distances into the ratio: \( \frac{\sqrt{(x-1)^2 + y^2}}{\sqrt{(x+1)^2 + y^2}} = \frac{1}{2} \). Square both sides to simplify: \( \frac{(x-1)^2 + y^2}{(x+1)^2 + y^2} = \frac{1}{4} \).

Step 4: Cross Multiply and Simplify

Cross multiplying gives: \( 4((x-1)^2 + y^2) = (x+1)^2 + y^2 \). Expand both sides: \( 4(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2 \). Simplify to get \(4x^2 - 8x + 4 + 4y^2 = x^2 + 2x + 1 + y^2 \).

Step 5: Move All Terms to One Side and Simplify

Reorganize the equation: \( 3x^2 - 10x + 3y^2 + 3 = 0 \). We see this is a circle equation, centered at \( x = \frac{10}{6} = \frac{5}{3} \), and y=0.

Step 6: Identify the Center of the Circle

The center of the identified circle is potentially the circumcenter because all points \( (x,y) \) where \( x = \frac{5}{3} \) and \( y=0 \) satisfy the ratio condition.

Key Concepts

Distance Formula

The distance formula is a key concept in coordinate geometry. It allows us to calculate the distance between two points in a coordinate plane. When you have two points, such as \((x_1, y_1)\) and \((x_2, y_2)\), the distance between them is found using the formula:\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For example, to find the distance from a point \( P(x,y) \) to \((1,0)\), you would substitute these into the formula like this: \( \sqrt{(x-1)^2 + y^2} \).
Using the distance formula is essential for solving various geometry problems, such as finding lengths and relationships between different points on the plane. Its value comes from transforming geometric concepts into algebraic expressions that can be easily solved.

Ratio of Distances

The concept of the ratio of distances involves comparing two separate distances. This comparison reveals meaningful geometric relationships between the points. In our problem, we are given that the ratio of the distance of any point \(P\) from one specific point \((1,0)\) to another point \((-1,0)\) is \(\frac{1}{2}\).
This means that if the distance from \(P\) to \((1,0)\) is calculated as \(d_1\), and the distance from \(P\) to \((-1,0)\) as \(d_2\), then \(\frac{d_1}{d_2} = \frac{1}{2}\). This ratio implies that the point \(P\) is closer to \((1,0)\) by a factor of \(2\) than it is to \((-1,0)\).
Applying the ratio effectively helps us understand spatial relationships and derive necessary conditions for other geometric constructs, like circles or triangles, in the coordinate plane.

Circumcenter

A circumcenter is a point that represents the heart of a circle passing through the three vertices of a triangle. This special point is equidistant from all three vertices of the triangle. In our context, if points \(A\), \(B\), and \(C\) lie on a circle due to all being equidistant from a common point, that point is the circumcenter.
Finding the circumcenter involves computations and understanding the layout of the triangle within the coordinate plane. Specifically, when given certain conditions such as distance ratios or centered geometric figures like a circle, it helps locate this central point.

• It is crucial in triangle geometry.
• Equidistant property relates to balanced distances.
• Use symmetrical properties for calculation.

In this problem, simplifying the given conditions eventually led us to identify the circumcenter, suggesting spatial equilibrium within the triangle formed by \(A\), \(B\), and \(C\).

Circle Equation

A circle equation describes all the points that make up a circle on the coordinate plane. It's powerful because once identified, it reveals critical properties like the center and radius of the circle.
The general form of a circle's equation is:\[ (x - h)^2 + (y - k)^2 = r^2 \]where \((h, k)\) is the center of the circle, and \(r\) is the radius.
In our exercise, solving the equation \(3x^2 - 10x + 3y^2 + 3 = 0\) revealed a circle. This can be transformed by completing the square into the standard circle form, though here, recognizing its pattern led us to identify the relationship between points and the circumcenter.

• Turning to the circle equation allows systematic approach.
• Once parameters \((h, k, r)\) are understood, geometric properties follow.
• Equations show spatial relations visibly in algebraic form.

Bringing all these elements together helps not only solve the problem but appreciate the geometry hidden in algebraic expressions.