Problem 50 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(L_{1}\) be a straight line passing through the origin and \(L_{2}\) be the straight line \(x+y=1\). If the interecpts made by the circle \(x^{2}+y^{2}-x+3 y=0\) on \(L_{1}\) and \(L_{2}\) are equal, then which of the following equations can represent \(L_{1} ?\) (A) \(x+y=0\) (B) \(x-y=0\) (C) \(7 y+2 x=0\) (D) \(x-7 y=0\)

Step-by-Step Solution

Answer

The line \(L_1\) is represented by option (B): \(x-y=0\).

Step 1: Intercepts of Circle on Line L2

First, we find the points of intersection of the circle \(x^2 + y^2 - x + 3y = 0\) with the line \(L_2: x + y = 1\). Substitute \(y = 1 - x\) into the circle's equation. This gives: \(x^2 + (1-x)^2 - x + 3(1-x) = 0\). Simplify to solve for \(x\).

Step 2: Solve Quadratic for x

Simplify the equation: \(x^2 + 1 - 2x + x^2 - x + 3 - 3x = 0\) leading to \(2x^2 - 6x + 4 = 0\). Divide the whole equation by 2 giving: \(x^2 - 3x + 2 = 0\). Factorize to get \((x-1)(x-2) = 0\). Hence, \(x = 1\) or \(x = 2\).

Step 3: Corresponding y Values

For \(x=1\), \(y = 0\) and for \(x = 2\), \(y = -1\). Thus, the intercepts on \(L_2\) are at points \((1, 0)\) and \((2, -1)\). The segment length is \(\sqrt{(2-1)^2 + (-1-0)^2} = \sqrt{2}\).

Step 4: Intercepts of Circle on Line L1

For the line \(L_1\) through the origin with intercepts equal to \(L_2\): Substitute \(y = mx\) into circle's equation: \(x^2 + (mx)^2 - x + 3(mx) = 0\). Simplify to \((1+m^2)x^2 - x(1-3m) = 0\).

Step 5: Factorize and Solve for x

The equation becomes \(x[(1+m^2)x - (1-3m)] = 0\). Thus, \(x = 0\) or \((1+m^2)x = 1-3m\). Use quadratic roots to find a consistent value of \(m\).

Step 6: Equating Intercepts

Given intercept lengths must be equal, solve \((1+m^2) \cdot \sqrt{2} = (1-3m)\). Simplify until a quadratic in terms of \(m\). By comparing magnitudes directly from possible equations, check when expressions equalize.

Step 7: Evaluate Given Options

Replace \(m\) based on given options: (A) \(m=-1\), (B) \(m=1\), (C) \(m=-\frac{2}{7}\), (D) \(m=\frac{1}{7}\). Calculate intercept relations and verify. Option (B) satisfies as both intercept lengths match.

Key Concepts

Circle Intercepts

In analytic geometry, understanding where a circle intercepts a straight line involves calculating the points at which the circle and the line intersect. This is a fundamental task when dealing with circles and lines on the coordinate plane.

To find the intercepts of a circle on a given line, you substitute the line's equation into the circle's equation. Take the equation of a circle, for example, like the one in the exercise:

• Circle: \[x^2 + y^2 - x + 3y = 0\]
• Line: \[x + y = 1\]

By substituting the line's equation into the circle's, we simplify and solve for the variable of interest. In our example, substituting \(y = 1 - x\) into the circle's equation gives a new quadratic equation in terms of \(x\).

Solving this equation yields the x-values of the intercept points, which can then be used to find the corresponding y-values. The intercepts of a circle with a line don't only tell us where they meet but also help in understanding various geometrical properties such as symmetry and the circle's behavior relative to the line.

Quadratic Equations

Quadratic equations are crucial when working with circle intercepts because they frequently arise when you set a line equation inside a circle's equation. The circle's equation becomes a quadratic after substituting the line's equation.

A quadratic equation generally takes the form: \[ax^2 + bx + c = 0\]To solve it, you'll first need to simplify it, possibly even divide through by common factors to make the equation more manageable, as shown in the example:\[2x^2 - 6x + 4 = 0 \]becomes \[x^2 - 3x + 2 = 0\]after dividing everything by 2.

Next, you can factor the quadratic if possible, like so: \[(x-1)(x-2) = 0\]This factorized form reveals the solutions \(x = 1\) and \(x = 2\), which help identify the points of intercept. Quadratics thus help in narrowing down specific points of interest such as intercepts or roots, crucial for understanding the geometry of the problem.

Line Equations

The equation of a line in analytic geometry can take various forms, such as slope-intercept form \(y = mx + c\) or the standard form \(ax + by + c = 0\). Each form provides a different perspective about the line.

In this problem, the line \(L_2\) was given by \(x + y = 1\), which describes a line with a slope \(-1\) and an intercept at \(0,1\). The line \(L_1\), on the other hand, needed to be determined based on the condition that its intercepts with the circle are equal to that of \(L_2\).

By substituting the line equations into the circle's equation, you can derive relationships that might appear quadratic, such as when determining intercept lengths via the formula: \[y = mx\]and substituting in \(L_1\)'s parameters. Exploring different values of \(m\) helps to find which line equation equates the intercepts correctly. For this problem, trying different given line options such as \(x-y=0\) helps to find the line that satisfies the condition: equal intercept lengths. The intercepts tell us so much more about a line including its interaction with other geometric shapes like circles or other lines.