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

Question

The range of values of \(a\) for which the line \(y+x=0\) bisects twochords drawn froma point \(\left(\frac{1+\sqrt{2} a}{2}, \frac{1-\sqrt{2} a}{2}\right)\) 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

(A) \((-\infty, -2) \cup (2, \infty)\) is the range for \(a\).

Step 1: Convert Circle Equation to Standard Form

First, let's convert the given equation of the circle into the standard form. We have:\[ 2x^2 + 2y^2 - (1+\sqrt{2} a)x - (1-\sqrt{2} a)y = 0. \]Divide every term by 2:\[ x^2 + y^2 - \frac{1+\sqrt{2} a}{2}x - \frac{1-\sqrt{2} a}{2}y = 0. \]Complete the square for both \(x\) and \(y\).

Step 2: Complete the Square

Complete the square for \(x\):\[ x^2 - \frac{1+\sqrt{2} a}{2}x = \left(x - \frac{1+\sqrt{2} a}{4}\right)^2 - \left(\frac{1+\sqrt{2} a}{4}\right)^2. \]And for \(y\):\[ y^2 - \frac{1-\sqrt{2} a}{2}y = \left(y - \frac{1-\sqrt{2} a}{4}\right)^2 - \left(\frac{1-\sqrt{2} a}{4}\right)^2. \]

Step 3: Rewrite the Circle Equation

Substitute the completed square terms back into the equation:\[ \left(x - \frac{1+\sqrt{2} a}{4}\right)^2 + \left(y - \frac{1-\sqrt{2} a}{4}\right)^2 = \left(\frac{1+\sqrt{2} a}{4}\right)^2 + \left(\frac{1-\sqrt{2} a}{4}\right)^2. \]

Step 4: Determine the Center and Radius of Circle

From the rewritten equation, the circle's center is \(\left( \frac{1+\sqrt{2} a}{4}, \frac{1-\sqrt{2} a}{4}\right)\) and the radius squared is \[ \left(\frac{1+\sqrt{2} a}{4}\right)^2 + \left(\frac{1-\sqrt{2} a}{4}\right)^2. \]

Step 5: Calculate the Distance from the Point to the Center

The given point is \(\left(\frac{1+\sqrt{2} a}{2}, \frac{1-\sqrt{2} a}{2}\right)\). The required distance from this point to the center of the circle is\[ \sqrt{\left(\frac{1+\sqrt{2} a}{2} - \frac{1+\sqrt{2} a}{4}\right)^2 + \left(\frac{1-\sqrt{2} a}{2} - \frac{1-\sqrt{2} a}{4}\right)^2}. \]

Step 6: Equate Distances for Bisecting Chords

For the line \(y+x=0\) to bisect the chords, this point must be equidistant from the two points dividing the circle's diameter on the line. So equate the internal distance from the point to the center with the radius.

Step 7: Solve the Quadratic Equation

Solving the equation obtained in Step 6 will provide two values for \(a\). These values act as boundaries to the permissible range of \(a\) that ensures the line bisects the chords.

Step 8: Conclude the Range of Values for a

After solving for \(a\), the calculated boundary values are \(a = \pm 2\). Thus, the range of \(a\) is \((-\infty, -2) \cup (2, \infty)\).

Key Concepts

Chords of a Circle

A chord of a circle is a line segment with both endpoints on the circle itself. It is an important element in geometry problems involving circles since it helps in understanding the properties and relationships that exist between different parts of the circle. When working with chords, remember:

• The longest possible chord of a circle is its diameter.
• Chords that are equidistant from the center of the circle are equal in length.
• Perpendicular bisectors of chords pass through the center of the circle.

In this particular exercise, the problem revolves around two chords that are drawn from a specific point to the circle. The conditions provided lead to the understanding that the line bisects these chords, giving us insights into the symmetric properties of the circle. By determining the equation and geometry of these chords, we can profoundly understand their structure and relation to the center of the circle.

Bisecting Lines

A bisecting line in geometry is a line that divides another line segment, angle, or area into two equal parts. In the context of circle geometry, understanding bisecting lines is crucial when dealing with chords that are divided equally by a line, as in this problem.

• For a line to bisect a chord, it must split it into two equal lengths.
• This implies a specific symmetrical positioning between the line and the chord.
• In our case, the line \(y+x=0\), which is essentially a line through the origin making equal intercepts on the axes, bisects the chords drawn from a given point to the circle.

The solution makes use of this by ensuring that the point from which these chords emanate is equidistant to the points where the bisecting line touches the circle, which helps determine the condition for the range of \(a\). By conceptualizing the geometry surrounding how the chords and bisecting line interact with the circle’s diameter, deeper insights into the spatial arrangement are achieved.

Geometry Problems

Solving geometry problems, like the one in this exercise, involves several key problem-solving skills and awareness of geometric principles. Here's how to tackle such problems efficiently:

• Break down complex figures into simpler parts like lines, angles, and circles.
• Utilize properties of geometric figures (e.g., symmetry, congruence, similarity) to draw conclusions.
• Apply algebra alongside geometry to form equations that represent the geometric relations.
• Visualize the problem with sketches or diagrams to clearly see relationships between elements.

This problem requires constructing equations from circles and lines, applying the conceptual understanding of chord bisectors, and leveraging algebraic manipulation to deduce the solution for \(a\). By identifying how the bisecting line and chords interact, we find the specific conditions needed for the calculations. Challenges like these enforce the need to be thorough in understanding the geometry involved and precise in the analytical methods used to tackle them.