Problem 67 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the line \(y=m x+1\) meets the circle \(x^{2}+y^{2}+3 x=0\) in two points equidistant from and on opposite sides of \(x\)-axis, then (a) \(3 m+2=0\) (b) \(3 m-2=0\) (c) \(2 m+3=0\) (d) \(2 m-3=0\)

Step-by-Step Solution

Answer

The correct condition is \(2m + 3 = 0\), so the answer is (c).

Step 1: Substitute the Line Equation into the Circle Equation

Substitute the line equation \( y = mx + 1 \) into the circle equation \( x^2 + y^2 + 3x = 0 \). This gives:\[ x^2 + (mx + 1)^2 + 3x = 0. \] Expanding this, we have:\[ x^2 + m^2x^2 + 2mx + 1 + 3x = 0. \] Group the \(x^2\) terms, the \(x\) terms, and the constants together:\[ (1 + m^2)x^2 + (2m + 3)x + 1 = 0. \]

Step 2: Determine the Condition for Symmetrical Intersection Points

Since the intersection points are equidistant from and on opposite sides of the \(x\)-axis, the roots of the quadratic equation must be equal in magnitude and opposite in sign. This means that the linear term of the quadratic equation must be zero. Therefore:\[ 2m + 3 = 0. \]

Step 3: Solve for \(m\)

From the condition \(2m + 3 = 0\), solve for \(m\):\[ 2m + 3 = 0 \] implies \( 2m = -3 \), so \( m = -\frac{3}{2} \).

Key Concepts

Intersection of Line and Circle

Understanding how a line intersects a circle is key in analytical geometry. This reveals where the two shapes meet on a graph. In our example, we determine the meeting points of the line represented by the equation \( y = mx + 1 \) with the circle described by \( x^2 + y^2 + 3x = 0 \).
To find the points of intersection, it's necessary to substitute the straight line equation into the circle's equation. This substitution helps in creating a new equation involving a single variable. Solving this equation tells us where the circle and the line cross each other.
Intersection points can be:

• Two distinct points
• One point (line is tangent to the circle)
• No points at all (line does not intersect the circle)

In this exercise, we are specifically looking at two intersection points.

Quadratic Equation

A quadratic equation is a second-degree polynomial typically expressed as \( ax^2 + bx + c = 0 \). The highest degree of the variable is two, which means the graph of the equation is a parabola. Understanding how to manipulate and solve quadratic equations is essential for finding the intersection of lines and circles, as they form as an outcome of the substitution of line and circle equations.
In the exercise, the resulting equation after substitution is \( (1 + m^2)x^2 + (2m + 3)x + 1 = 0 \), which is a quadratic form. This equation can be solved using various methods like:

• Factoring
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Completing the square

The choice of method depends on the specifics of the equation. However, in this particular problem, understanding symmetry is key.

Symmetrical Intersection Points

In some cases, two intersection points can be symmetrical about an axis. For this exercise, the points are equidistant from and on opposite sides of the \( x \)-axis. This implies certain mathematical conditions. For symmetrical points relative to the \( x \)-axis, the linear coefficient in the quadratic equation, \((2m + 3)x\), must be zero.
When the linear term \( b = 0 \) in \( ax^2 + bx + c = 0 \), it indicates the roots (or \( x \)-values of intersection points) are opposite in sign but equal in magnitude. This stems from the roots being symmetric with respect to the origin.
The exercise uses this concept to derive that \( 2m + 3 = 0 \). Solving this equation gives us \( m = -\frac{3}{2} \) as the condition required for symmetry in the intersection points. This condition guarantees that the placements of the intersection points match the problem's symmetry requirement.