Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The range of values of \(a\) for which the line \(y+x=0\) bisectstwochords drawn fromapoint \(\left(\frac{1+\sqrt{2} a}{2}, \frac{1-\sqrt{2} a}{2}\right)\) to the circle \(2 x^{2}+2 y^{2}-(1+\sqrt{2} a) x-(1-\sqrt{2} a) y=0\) is (A) \((-\infty,-2) \cup(2, \infty)\) (B) \((-2,2)\) (C) \((2, \infty)\) (D) none of these

Step-by-Step Solution

Answer

The range of values of \(a\) is \((-\infty,-2) \cup (2,\infty)\).

Step 1: Simplify the Circle Equation

The given equation of the circle is \(2x^2 + 2y^2 - (1+\sqrt{2}a)x - (1-\sqrt{2}a)y = 0\). Divide throughout by 2 to simplify: \(x^2 + y^2 - \frac{(1+\sqrt{2}a)}{2}x - \frac{(1-\sqrt{2}a)}{2}y = 0\).

Step 2: Find the Center and Radius of the Circle

Complete the square for the terms in \(x\) and \(y\): \((x - \frac{1+\sqrt{2}a}{4})^2 + (y - \frac{1-\sqrt{2}a}{4})^2 = \left(\frac{\sqrt{3+a^2}}{4}\right)^2\). The center is at \((\frac{1+\sqrt{2}a}{4}, \frac{1-\sqrt{2}a}{4})\) and the radius is \(\frac{\sqrt{3+a^2}}{4}\).

Step 3: Understand Condition for Line Bisecting Chords

For the line \(y + x = 0\) to bisect chords drawn from point \(\left(\frac{1+\sqrt{2}a}{2}, \frac{1-\sqrt{2}a}{2}\right)\), it must also pass through the center of the circle. Thus, these need to be collinear.

Step 4: Derive the Condition for Collinearity

The point \(\left(\frac{1+\sqrt{2}a}{2}, \frac{1-\sqrt{2}a}{2}\right)\), the center \(\left(\frac{1+\sqrt{2}a}{4}, \frac{1-\sqrt{2}a}{4}\right)\) and the line \(y + x = 0\) implies that these points satisfy \(y + x = 0\), resulting in: \(\frac{1+\sqrt{2}a}{4} + \frac{1-\sqrt{2}a}{4} = 0\), simplifying to \(a = \pm 2\).

Step 5: Determine Range for Parameter a

The condition derived implies the boundaries \(a = \pm 2\) are not part of the solution; hence, the values of \(a\) for which the line bisects are \((-\infty, -2) \cup (2, \infty)\).

Key Concepts

Circle Equation

To begin understanding analytical geometry, we need to grasp the concept of the circle equation. A circle can be described in a standard form equation such as \[(x - h)^2 + (y - k)^2 = r^2\]where

• \((h, k)\) represents the center of the circle,
• \(r\) is the radius.

For example, in the problem provided, the given and simplified circle equation is: \[x^2 + y^2 - \frac{(1 + \sqrt{2}a)}{2}x - \frac{(1 - \sqrt{2}a)}{2}y = 0\]By completing the square, we can identify that the circle's center is at \(\left(\frac{1+\sqrt{2}a}{4}, \frac{1-\sqrt{2}a}{4}\right)\),and the radius is \(\frac{\sqrt{3+a^2}}{4}\).This process of completing the square is key for restructuring the equation into the standard form, aiding in identifying essential circle properties like the center and radius.

Line Bisecting Chords

In geometry, a line is said to bisect a chord if it divides the chord into two equal segments. In context with our problem, the line \(y + x = 0\)should be able to bisect chords of the circle drawn from a specific point. What makes this problem challenging is that for the line to bisect chords, it should pass through the circle's center.
This requirement helps establish conditions for the possible range for parameter \(a\).When the line passes through the center of the circle, it ensures symmetry which is a geometric prerequisite for bisecting the chords evenly. Understanding this helps us infer that the relationship between line and circle parameters is crucial in determining collinearity and other geometrical properties.

Collinearity

Collinearity is a condition where points lie on the same straight line. In this exercise, identifying collinearity helps us relate certain points with the line equation. The points involved are:

• The given point from which chords are drawn, \(\left(\frac{1+\sqrt{2}a}{2}, \frac{1-\sqrt{2}a}{2}\right)\),
• The center of the circle, \(\left(\frac{1+\sqrt{2}a}{4}, \frac{1-\sqrt{2}a}{4}\right)\),

For them to be collinear with the line \(y + x = 0\),both sets of coordinates must satisfy the line's equation. This translates into an equation \(\frac{1+\sqrt{2}a}{4} + \frac{1-\sqrt{2}a}{4} = 0\)which simplifies to \(a = \pm 2\).Identifying this collinear condition helps us in calculating permissible values for \(a\), establishing a connection between geometry and algebra.

Range of Parameter

Determining the range of a parameter involves understanding which values it can take while still satisfying certain equations or inequalities. In this problem, after establishing that \(a = \pm 2\)are boundary values where the collinear condition holds, it means that the actual range where \(a\)'s condition is met but doesn't split into the defined limits.
Therefore, the values satisfying the bisecting condition for the chords of the circle are prevented from including these particular points \((a = \pm 2)\).The range of \(a\)in the problem is then \((-\infty, -2) \cup (2, \infty)\). This means \(a\)can take any real number except \(a = \pm 2\). This captures the essence of restrictions based on geometric conditions leading to analytical outcomes.