Problem 52 - A Complete Resource Book in Mathematics for JEE Main

Question

A point moves such that the sum of the squares of its distances from the sides of a square of side unity is equal to 9 . The locus of the point is a circle such that (A) centre of the circle coincides with that of square (B) centre of the circle is \(\left(\frac{1}{2}, \frac{1}{2}\right)\) (C) radius of the circle is 2 (D) all the above are true

Step-by-Step Solution

Answer

All given options (A, B, C) are true.

Step 1: Understanding the Problem

We need to find the locus of a point such that the sum of the squares of its distances from the sides of a unit square equals 9. This locus turns out to be a circle, and we have to determine the center and radius of this circle.

Step 2: Identify Coordinates of Square

Consider a unit square with its sides along the x-axis and y-axis, having vertices at (0,0), (1,0), (1,1), and (0,1). This means the square is centered at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Step 3: Express Sum of Squared Distances

For a point \((x, y)\), the distances to the sides are \(x\), \(y\), \(1-x\), and \(1-y\). The sum of the squares of these distances is: \(x^2 + y^2 + (1-x)^2 + (1-y)^2 = 9\).

Step 4: Simplify the Expression

Simplify the expression: \(x^2 + y^2 + (1-x)^2 + (1-y)^2 = 9\) becomes \(x^2 + y^2 + 1 - 2x + x^2 + 1 - 2y + y^2 = 9\) which simplifies to \(2x^2 + 2y^2 - 2x - 2y + 2 = 9\).

Step 5: Rearrange and Complete the Square

Rearrange and simplify: \(2(x^2 - x) + 2(y^2 - y) = 7\) or \(x^2 - x + y^2 - y = \frac{7}{2}\). Complete the square for \(x\) and \(y\): \((x - \frac{1}{2})^2 - \frac{1}{4} + (y - \frac{1}{2})^2 - \frac{1}{4} = \frac{7}{2}\).

Step 6: Set the Equation of the Circle

Combine terms: \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \frac{7}{2} + \frac{1}{2}\). This results in \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\).

Step 7: Extract Circle's Properties

The equation of the circle \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\) has its center at \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius 2.

Key Concepts

Circle Properties

The concept of a circle is central in understanding the locus of a point in geometric problems. Circles have distinctive properties, one of which is their set of points equidistant from a center point. In this exercise, the circle is described by an equation derived from the given geometric conditions.

When you have a circle, the key properties to examine are:

• Center: This is the fixed point from which every point on the circle is the same distance. Here, the circle's center is at \( \left(\frac{1}{2}, \frac{1}{2}\right) \), coinciding with the center of the square.
• Radius: The constant distance from the center to any point on the circle. In this problem, the radius is 2 units.
• Equation of the circle: Typically written in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Understanding these properties helps to visualize and solve problems like finding the locus of a point, which is often a circle.

Completing the Square

Completing the square is a mathematical technique used to rewrite quadratic expressions, making it easier to identify features such as vertex form for parabolas or standard form for circles. It transforms expressions for easier manipulation and solution finding.

In this exercise, completing the square involved:

• Identifying Quadratic Terms: Here, terms like \(x^2 - x\) and \(y^2 - y\) were selected for transformation.
• Rewriting Terms: To transform these into perfect squares, half of the linear coefficient is squared and added/subtracted inside the equation. This changes \(x^2 - x\) to \((x - \frac{1}{2})^2 - \frac{1}{4}\) and similarly for \(y\).
• Balancing the Equation: After rearranging terms into perfect squares, ensure the equation remains balanced by adjusting the constant terms. This culminates in the equation \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\).

This step is crucial as it provides the means to convert the quadratic equation into the recognizable circle equation form.

Equation of a Circle

The equation of a circle provides a concise way to represent all the points forming a circle on a coordinate plane. Derived from the Pythagorean theorem, it ensures that the distance from any point on the circle to its center is constant, known as the radius.

The general formula for 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 the exercise,** the completion of the square gives the circle's equation as \((x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = 4\). This form directly reveals:

• Center: \(\left(\frac{1}{2}, \frac{1}{2}\right)\), which is central to understanding its orientation within the square.
• Radius: \(2\), derived from \(\sqrt{4}\), dictating the circle's size.

Using the circle equation is beneficial for calculating and confirming properties like radius and center, which are crucial for solving geometric problems involving circle surfaces and intersections.