Problem 4 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the line \((y-2)=m(x+1)\) intersects the circle \(x^{2}+\) \(y^{2}+2 x-4 y-3=0\) at two real distinct points, then the number of possible values of \(m\) is (A) 2 (B) 1 (C) any real value of \(m\) (D) none of these

Step-by-Step Solution

Answer

C) Any real value of m

Step 1: Standardize the Line Equation

The given line equation is \( y - 2 = m(x + 1) \). To standardize this, solve for \( y \):\[ y = mx + m + 2. \]

Step 2: Standardize the Circle Equation

The given circle equation is \( x^2 + y^2 + 2x - 4y - 3 = 0 \). Rewrite it by completing the square for both \( x \) and \( y \).\[ (x+1)^2 - 1 + (y-2)^2 - 4 = 3 \]which simplifies to:\[ (x+1)^2 + (y-2)^2 = 8. \]This represents a circle centered at \((-1, 2)\) with radius \( \sqrt{8} \).

Step 3: Substitute Line Equation into Circle Equation

Substitute \( y = mx + m + 2 \) from Step 1 into the circle equation \((x+1)^2 + (y-2)^2 = 8\).\[ (x+1)^2 + ((mx + m + 2) - 2)^2 = 8 \]which simplifies to \( (x+1)^2 + (mx + m)^2 = 8 \).

Step 4: Set Up Quadratic Equation in Terms of \( x \)

Expanding the equation from Step 3,\[ (x+1)^2 = x^2 + 2x + 1 \]and\[ (mx + m)^2 = m^2x^2 + 2m^2x + m^2. \]Combine these to form one equation:\[ x^2 + 2x + 1 + m^2x^2 + 2m^2x + m^2 = 8. \]This forms:\[ (1+m^2)x^2 + (2+2m^2)x + (1+m^2-8) = 0. \]

Step 5: Analyze the Discriminant

For the line to intersect the circle at two real and distinct points, the discriminant of the quadratic \( ax^2 + bx + c = 0 \) must be positive. The discriminant \( \Delta \) is given by:\[ \Delta = b^2 - 4ac. \]Substitute the equation from Step 4, \(a = 1+m^2\), \(b = 2+2m^2\), and \(c = 1+m^2-8\), to get:\[ \Delta = (2+2m^2)^2 - 4(1+m^2)(1+m^2-8) > 0. \]

Step 6: Simplify the Discriminant

Simplify the discriminant expression:\[ \Delta = (2 + 2m^2)^2 - 4(1 + m^2)(m^2 - 7) > 0. \]Expanding gives:\[ 4 + 8m^2 + 4m^4 - 4m^4 + 28(1 + m^2) > 0. \]Thus, simplify and solve for m:\[ 8m^2 + 4 + 28m^2 + 28 > 0 \]This simplifies to:\[ 36m^2 + 32 > 0. \]

Step 7: Solve for Values of \(m\)

The inequality \( 36m^2 + 32 > 0 \) is always true for any real number \( m \) because \( 36m^2 > 0 \) for all \( m eq 0 \), therefore, the line always intersects the circle at two distinct points. Hence, any real value of \( m \) works.

Key Concepts

Circle Equation

A circle in a two-dimensional plane can be represented by an algebraic equation. The general form of the circle’s equation is
\[ (x - h)^2 + (y - k)^2 = r^2 \]where:

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

The given circle equation in the exercise is \( x^2 + y^2 + 2x - 4y - 3 = 0 \). To convert this into standard form, we need to complete the square for the terms concerning \( x \) and \( y \).
Completing the square involves rearranging the equation into a more recognizable form like this:

(a) The \( x^2 + 2x \) terms become \((x+1)^2 - 1\).

(b) The \( y^2 - 4y \) terms become \((y-2)^2 - 4\).

After completing these steps, the circle equation becomes:
\[ (x+1)^2 + (y-2)^2 = 8 \]
This tells us that the circle is centered at \((-1, 2)\) and its radius is \( \sqrt{8} \), which comes directly from interpreting the standard format.

Line Equation

Lines can also be represented by equations in a two-dimensional plane. The basic form of a line equation is the slope-intercept form given by:
\[ y = mx + c \]where:

• \( m \) represents the slope of the line, showing how steep the line rises or falls.
• \( c \) represents the y-intercept, which is where the line crosses the y-axis.

In the exercise, the line equation is initially given as \( y - 2 = m(x + 1) \). Solving for \( y \) gives us:
\[ y = mx + m + 2 \]
This shows that the slope \( m \) influences the angle at which the line intersects other curves or straight paths on the graph.
Understanding the role of \( m \) helps determine how the line behaves when interacting with features like a circle.

Discriminant

The discriminant is a key concept when working with quadratic equations, especially when considering the intersection of lines and circles. The quadratic equation takes the form:
\[ ax^2 + bx + c = 0 \]
The discriminant \( \Delta \) is calculated as:
\[ \Delta = b^2 - 4ac \]
The discriminant reveals the nature of the roots (solutions) of the quadratic equation:

• If \( \Delta > 0 \), the equation has two distinct real roots.
• If \( \Delta = 0 \), the equation has exactly one real root (the line is tangent to the circle).
• If \( \Delta < 0 \), there are no real roots (the line does not intersect the circle).

To find potential values of \( m \) for which the line intersects a circle at two distinct points, the discriminant of the quadratic formed by substituting the line equation into the circle’s equation must be greater than zero.
In the problem, this led to solving \( 36m^2 + 32 > 0 \), indicating that for any real value of \( m \), the discriminant is positive, ensuring two points of intersection.