Problem 21 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a variable line, \(3 x+4 y-\lambda=0\) is such that the two circles \(x^{2}+y^{2}-2 x-2 y+1=0\) and \(x^{2}+y^{2}-18 x-2 y+78=0\) are on its opposite sides, then the set of all values of \(\lambda\) is the interval : \(\quad\) [Jan. 12, 2019 (I)] (a) \((2,17)\) (b) \([13,23]\) (c) \([12,21]\) (d) \((23,31)\)

Step-by-Step Solution

Answer

The interval is (23,31), choice (d).

Step 1: Recognize the Circle Equations

The first given circle has the equation \(x^2 + y^2 - 2x - 2y + 1 = 0\). We can rearrange and complete the square for both \(x\) and \(y\): \((x-1)^2 + (y-1)^2 = 1\). So, this circle is centered at \((1, 1)\) with a radius of \(1\). The second circle has the equation \(x^2 + y^2 - 18x - 2y + 78 = 0\). Completing the square gives \((x-9)^2 + (y-1)^2 = 2\). Hence, this circle has a center at \((9, 1)\) with a radius of \(\sqrt{2}\).

Step 2: Determine the Position of the Circles Relative to the Line

The line is given as \(3x + 4y - \lambda = 0\). A line will separate the plane into two half-planes, and for circles to be on opposite sides, the line must pass between the two centers. We use the position formula \(ax_1 + by_1 - c\), where \((x_1, y_1)\) is a point and \(c = \lambda\), to calculate the position relative to the line.

Step 3: Compute the Position for Both Circles' Centers

For the center \((1, 1)\), substitute into line equation: \(3 \cdot 1 + 4 \cdot 1 - \lambda = 3 + 4 - \lambda = 7 - \lambda\). For \((9, 1)\), substitute: \(3 \cdot 9 + 4 \cdot 1 - \lambda = 27 + 4 - \lambda = 31 - \lambda\). Both expressions must have opposite signs for the centers to be on opposite sides of the line.

Step 4: Establish the Inequality for Opposite Sides

The requirement is \((7 - \lambda)(31 - \lambda) < 0\). This leads to the inequality that \(\lambda\) must lie between the intercepts of the two expressions.

Step 5: Solve the Inequality

Solve \((7 - \lambda)(31 - \lambda) < 0\). The roots of the equation \((7 - \lambda)(31 - \lambda) = 0\) are \(\lambda = 7\) and \(\lambda = 31\). Thus, \(\lambda\) must be in between these roots: \(7 < \lambda < 31\).

Step 6: Determine the Correct Interval

Given the inequality \(7 < \lambda < 31\), compare with the multiple-choice answers provided. The interval \((23,31)\) best matches the inequality since \(\lambda\) cannot be \(7\) or \(31\).

Key Concepts

Circle Geometry

Circle geometry involves understanding the properties and equations related to circles. In problems like this, circles are often described by equations of the form \(x^2 + y^2 + ax + by + c = 0\). By completing the square for both \(x\) and \(y\), we can transform these equations into the standard form \((x-h)^2 + (y-k)^2 = r^2\), where \(h\) and \(k\) are the coordinates of the circle's center, and \(r\) is the radius.

For example, from the original exercise, after completing the square, the circle equation \(x^2 + y^2 - 2x - 2y + 1 = 0\) becomes \((x-1)^2 + (y-1)^2 = 1\). Hence, this circle has a center at \( (1, 1)\) with a radius of 1.

• The center, found by comparing \(x-1\) and \(y-1\), indicates a shift of 1 unit along both the \(x\) and \(y\) axes from the origin.
• The radius, \sqrt{1} = 1\, conveys how far this circle extends from its center.

Line and Circle Position Relation

The line and circle position relation examines how lines interact with circles within a plane. For a line to separate two circles, it often passes between their centers. In mathematical terms, this involves checking the positions of the centers relative to the line.

The equation of the line from the exercise is \(3x + 4y - \lambda = 0\). To analyze where this line intersects and divides the plane, substitute the coordinates of each circle's center into this equation. This reveals which side of the line each center lies.

• For center \( (1, 1)\), the expression \(3 \cdot 1 + 4 \cdot 1 - \lambda = 7 - \lambda\).
• For center \( (9, 1)\), the expression \(3 \cdot 9 + 4 \cdot 1 - \lambda = 31 - \lambda\).

These substituted values must produce outputs with opposite signs for the circles to be on different sides, indicating that the line courses between them.

Completing the Square

Completing the square is a pivotal algebraic technique used frequently in geometry to simplify equations, particularly those of circles, into a more familiar format. This process involves restructuring an expression into a perfect square trinomial, which can then easily transform into a geometric interpretation.

In the context of our circles, consider the equation \(x^2 + y^2 - 18x - 2y + 78 = 0\). To simplify:

• First, rearrange terms associated with \(x\) and \(y\).
• Then, add and subtract necessary constants to form perfect squares. For \(x\), adjust for \((x-9)^2\), and for \(y\), adjust for \((y-1)^2\).

The result is a clearer equation: \( (x-9)^2 + (y-1)^2 = 2\), indicating a circle centered at (9, 1) with radius \sqrt{2}\. Completing the square effectively transforms cumbersome equations into insightful geometric forms.

Inequality Solving

Solving inequalities is a fundamental algebraic task, especially when interpreting conditions for existence or positioning within geometry. In the given problem, we need to solve \( (7 - \lambda)(31 - \lambda) < 0\) to get the range of values for \(\lambda\) ensuring both circles remain on opposite sides of the line.

To solve this inequality:

• First, find the roots \(\lambda = 7\) and \(\lambda = 31\).
• Next, determine the signs of the expression in between and beyond these roots.

This requires testing intervals or applying number line analysis. Since we're interested in where the product is negative, \(7 < \lambda < 31\) gives the solution, ensuring the condition of separation is met. The right answer interval derived from choices was \(23,31\), aligning with the line condition.