Problem 50 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(x - y = 0\).

Step 1: Understand the Equation of Circle

The given circle is represented by the equation \(x^2 + y^2 - x + 3y = 0\). We will rearrange this into the standard form of a circle equation \( (x - h)^2 + (y - k)^2 = r^2 \) to better understand its properties, such as center and radius.

Step 2: Transform Circle Equation

The equation \(x^2 + y^2 - x + 3y = 0\) can be written by completing the square for both \(x\) and \(y\). Completing the square, we have:\[(x - \frac{1}{2})^2 - \frac{1}{4} + (y + \frac{3}{2})^2 - \frac{9}{4} = 0 \]\[(x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{1}{4} + \frac{9}{4} \]\[(x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{10}{4} = \frac{5}{2} \]

Step 3: Recognize Equation of Lines

We have two lines: \(L_1\), passing through origin, can be represented as \(y = mx\) or any other homogeneous form, and \(L_2\) is explicitly given as \(x+y=1\). Our task is to compare the intercepts of line \(L_1\) and \(L_2\) with the circle.

Step 4: Calculate Intercepts on L2

For the line \(L_2: x+y=1\), substitute into the circle equation to find intercept points by solving for \(x\) and \(y\). Replacing \(y = 1-x\) in the circle equation, \\[ x^2 + (1-x)^2 - x + 3(1-x) = 0 \] \Expanding and simplifying, this results in quadratic in terms of \(x\): \\[ 2x^2 - 8x + 4 = 0 \] \Solving this quadratic equation gives the roots \(x = 2\), and substituting back \(y = 1-x\), results in identical intercepts.

Step 5: Compare with L1 Intercepting Points

Since intercepts are equal on both lines with the same radius from the circle, we substitute potential \(L_1\) values that satisfy \(ax + by = 0\) passing through the circle's center \((\frac{1}{2}, -\frac{3}{2})\). Only lines maintaining symmetrical path would satisfy equal intercepts.

Step 6: Verification of Line Options

Evaluate the options for \(L_1\): If intercepts on both lines are equal \**(A)** \(x + y = 0\): Equation implies symmetry on center and violates terms. \**(B)** \(x - y = 0\): Balanced intercept equality hence possible \(x=y\). \**(C)** \(7y + 2x = 0\): Check configurations on transformed circle equation. \**(D)** \(x - 7y = 0\): Unbalanced slope contradicts symmetry.

Step 7: Conclusion

Checking the conditions, one finds that the option \(x-y=0\) aligns with balanced algebra between intercepts showing the least resistance and adherence to rule \(m = 1\). Thus line \(L_1\) is valid with choice \(\boxed{B}.\)

Key Concepts

Circle Equation

A circle equation is an expression that represents all the points (x, y) located at a fixed distance (the radius) from a central point (the center) in a plane. The standard form of a circle equation is given by \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center of the circle and \(r\) is its radius.
To convert a general form like \(x^2 + y^2 - x + 3y = 0\) into the standard form, we use a method called "completing the square". This involves rearranging and transforming the equation to reveal the center and radius. In this case, completing the square gives us \((x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{5}{2}\). This circle is centered at \((\frac{1}{2}, -\frac{3}{2})\) with a radius of \(\sqrt{\frac{5}{2}}\).
Understanding the circle's equation helps us determine how it might interact with various lines, as lines can intersect the circle at specific points.

Line Equation

A line equation in its simplest form is expressed as \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept. For lines through the origin, this becomes \(y = mx\), or equivalently, \(ax + by = 0\). The line equation helps us understand how a line stretches across a plane and how it might intersect with other geometrical shapes.
In the given problem, line \(L_1\), which passes through the origin, can be represented by equations such as \(y = mx\). This allows us to explore various possibilities for \(L_1\) and check which aligns with the given circle and conditions of the problem.
Additionally, line \(L_2\) is explicitly given as \(x+y=1\). Here, we can see that it intersects the x-axis at \(1,0\) and the y-axis at \(0,1\), forming what's known as intercepts of the line.

Intercepts

Intercepts refer to the points where a line crosses the x-axis and y-axis. These are crucial in understanding the geometry of lines and circles, as they help in determining where the circle and the line might intersect each other.
For line \(L_2: x+y=1\), the intercepts are found by examining where the line crosses the axes. The x-intercept occurs where \(y = 0\), leading to \(x = 1\), and the y-intercept occurs where \(x = 0\), leading to \(y = 1\). These create intercepts at points \((1,0)\) and \((0,1)\).
When solving problems like this, knowing the intercepts on a circle can help determine equal lengths or symmetrical properties. For a circle such as \((x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{5}{2}\), it's essential that intercepts from \(L_1\) and \(L_2\) are the same for any given line \(L_1\).

Quadratic Equation

The quadratic equation often emerges when solving for intercepts between a circle and a line. This is because substituting the line's equation into the circle's equation reduces the problem to finding the roots of a quadratic equation, revealing where intersecting points occur.
In our exercise, when solving for the intercept of \(L_2\) with the circle, after substitution and simplification, we get the equation \(2x^2 - 8x + 4 = 0\). Solving this quadratic equation helps us find the intercepts, namely the values of \(x\) that make both the circle and the line equations true at the same time.
Solving a quadratic equation often involves using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which gives the points of intersection. In this context, it allows us to check if different potential lines for \(L_1\) will create similar points on the circle.