Problem 64 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If each of the lines \(5 x+8 y=13\) and \(4 x-y=3\) contains a diameter of the circle \(x^{2}+y^{2}-2\left(a^{2}-7 a+11\right) x-2\left(a^{2}-6 a+6\right) y+b^{3}+1=0\), then (a) \(a=5\) and \(b \notin(-1,1)\) (b) \(a=1\) and \(b \notin(-1,1)\) (c) \(a=2\) and \(b \notin(-\infty, 1)\) (d) \(a=5\) and \(b \in(-\infty, 1)\)

Step-by-Step Solution

Answer

The correct answer is (d) \(a=5\) and \(b \in(-\infty, 1)\).

Step 1: Equations of Lines in Diameter Form

To determine if the lines given contain a diameter of the circle, we start by remembering that if a line is a diameter of a circle, it must pass through the center of the circle. The equations of lines are given as \(5x + 8y = 13\) and \(4x - y = 3\).

Step 2: Find the Center of the Circle

The equation of the circle in standard form is \(x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^3 + 1 = 0\). The center \( (h, k) \) of the circle is derived from comparing this equation to \( (x - h)^2 + (y - k)^2 = r^2 \). Thus, \( h = a^2 - 7a + 11 \) and \( k = a^2 - 6a + 6 \).

Step 3: Substitute Center into Line Equations

Substitute \( x = a^2 - 7a + 11 \) and \( y = a^2 - 6a + 6 \) into the first line equation \(5x + 8y = 13\):\[5(a^2 - 7a + 11) + 8(a^2 - 6a + 6) = 13\] Solve this equation for \(a\).

Step 4: Solve for 'a' using First Line

Simplify and solve the first equation:\[5a^2 - 35a + 55 + 8a^2 - 48a + 48 = 13\]\[13a^2 - 83a + 103 = 13\]\[13a^2 - 83a + 90 = 0\]Use the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to further simplify:\(a = \frac{83 \pm \sqrt{83^2 - 4 \cdot 13 \cdot 90}}{2 \cdot 13}\)Calculate values: \(a = 5\) or \(a = 1\).

Step 5: Substitute Center into Second Line

Substitute \( x = a^2 - 7a + 11 \) and \( y = a^2 - 6a + 6 \) into the second line equation \(4x - y = 3\):\[4(a^2 - 7a + 11) - (a^2 - 6a + 6) = 3\] Solve this equation for \(a\).

Step 6: Solve for 'a' using Second Line

Simplify and solve the second equation:\[4a^2 - 28a + 44 - a^2 + 6a - 6 = 3\]\[3a^2 - 22a + 38 = 3\]\[3a^2 - 22a + 35 = 0\]Use the quadratic formula:\(a = \frac{22 \pm \sqrt{22^2 - 4 \cdot 3 \cdot 35}}{2 \cdot 3}\)Calculate values: \(a = 5\) or \(a = 1\).

Step 7: Determine Valid 'a' and Solve for 'b'

Since both lines are diameters, we consider values obtained from both steps: \(a = 5\) and \(a = 1\). These match both equations. To find \(b\), note that it must satisfy \(b otin (-1,1)\). Options for \(b^3 + 1\) conditions steer towards \(-1\). For non-polynomials, solve for \(b\) such that \(b otin (-1, 1)\). If \(a = 5\), \(b otin (-\infty, 1)\) is correct.

Key Concepts

Equations of Lines

Understanding the equations of lines is crucial in grasping this problem. Lines can be represented in the form of a linear equation like the two provided: \(5x + 8y = 13\) and \(4x - y = 3\). These equations represent straight lines on a coordinate plane.
The coefficients of \(x\) and \(y\) determine the slope of the line, while the constant term indicates where the line intersects the y-axis.
The line equations tell us the relationship between the variables \(x\) and \(y\), which we use to determine if these lines can serve as diameters of the given circle.

Circle Geometry

Circle geometry involves understanding how lines and other shapes relate to circles. A circle is defined by all the points equidistant from a central point, known as the center.
The given equation \(x^2 + y^2 - 2(a^2 - 7a + 11)x - 2(a^2 - 6a + 6)y + b^3 + 1 = 0\) represents a circle. By transforming it into the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\), we can identify the center \((h, k)\) and potentially, the radius \(r\).
Understanding these relationships helps us determine if a line is a diameter, which is possible if the line passes through the circle's center.

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \(ax^2 + bx + c = 0\). When trying to find the variable \(a\) in this exercise, we perform an initial simplification which leads to solving two quadratic equations:
\[13a^2 - 83a + 90 = 0\]
and
\[3a^2 - 22a + 35 = 0\].
The formula is given by \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). By plugging the respective coefficients into this formula, we identify potential values for \(a\), which in this case are \(a = 5\) and \(a = 1\).
This kind of calculation shows how the quadratic formula can provide solutions for the problem given certain constraints.

Center of Circle

In mathematics, the center of a circle is a pivotal concept, particularly for tasks involving diameters or tangents. In this problem, the center \((h, k)\) is calculated using transformations of the circle's general equation.
The transformation shows \(h = a^2 - 7a + 11\) and \(k = a^2 - 6a + 6\). This extraction process highlights the key steps to find the center, especially when the equation is not initially in its simplest form.
Knowing this center is essential because any diameter of the circle must pass through this point, helping us solve for other unknowns, such as the variable \(b\) once \(a\) is determined.