Problem 116 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is (A) \(x^{2}+y^{2}-2 x+2 y-23=0\) (B) \(x^{2}+y^{2}-2 x-2 y-23=0\) (C) \(x^{2}+y^{2}+2 x+2 y-23=0\) (D) \(x^{2}+y^{2}+2 x-2 y-23=0\)

Step-by-Step Solution

Answer

The equation of the circle is (A) \(x^2 + y^2 - 2x + 2y - 23 = 0\).

Step 1: Determine the Center of the Circle

The given lines are diameters of the circle and are perpendicular, as lines of a circle that intersect at 90 degrees form the diameters. To find the point of intersection of the lines, solve the system of equations:\[\begin{align*}2x + 3y + 1 &= 0 \3x - y - 4 &= 0\end{align*}\]Substitute or eliminate one variable to solve for the other. Solving gives us the center \((1, -1)\).

Step 2: Determine the Radius of the Circle

A circle's circumference is given as \(10\pi\). Since the circumference \(C = 2\pi r\), equate and solve for \(r\):\[2\pi r = 10\pi\]\[r = 5\]Thus, the radius \(r\) is 5.

Step 3: Write the Equation of the Circle

With the center \((h, k) = (1, -1)\) and the radius \(r = 5\), the general equation of a circle is:\[(x - h)^2 + (y - k)^2 = r^2\]Substituting, we get:\[(x - 1)^2 + (y + 1)^2 = 25\]Expanding, this equation becomes:\[x^2 - 2x + 1 + y^2 + 2y + 1 = 25\]\[x^2 + y^2 - 2x + 2y - 23 = 0\]

Step 4: Identify the Correct Option

Compare the expanded equation \(x^2 + y^2 - 2x + 2y - 23 = 0\) with the given options.The equation matches option (A):\[x^2 + y^2 - 2x + 2y - 23 = 0\]

Key Concepts

Geometry

In geometry, circles are fascinating shapes characterized by their constant distance from a center point. This distance is known as the radius. When dealing with problems involving circles, there are a few geometric concepts to keep in mind:

• A circle can be uniquely defined by its center and radius.
• The diameter is double the radius, stretching across the entire circle through the center.
• Any line that touches the circle at exactly one point is called a tangent.
• The circumference is the total distance around the circle.

The concept of diameters intersecting perpendicularly forms an interesting geometrical idea. Here, since the lines lie along the diameters, they intersect orthogonally at the center of the circle.

System of Linear Equations

A system of linear equations is a set of two or more equations involving two or more variables. Solving these systems of equations helps us find the point where the conditions of the equations meet. In this exercise:

• We have the equations: \(2x + 3y + 1 = 0\) and \(3x - y - 4 = 0\).
• The objective is to find the intersection point, which represents the center of the circle.
• This is achieved by either substitution or elimination methods, designed to reduce the system to a single equation in one variable.

By solving this system, we determined that the intersection point, or the center of the circle, is \((1, -1)\). This knowledge is crucial for defining the circle's equation.

Circle Properties

Circle properties are essential when determining the equation of a circle. The equation is pivotal for analyzing its geometry and position in the coordinate plane. The general equation of a circle 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.

Given our problem, we found:

• Center \((h, k) = (1, -1)\)
• Radius \(r = 5\), calculated from the circumference using the formula \(C = 2\pi r\).

Thus, substituting these values into the circle's equation, we derived: \[(x - 1)^2 + (y + 1)^2 = 25\]. Expanding this confirmed the equation \(x^2 + y^2 - 2x + 2y - 23 = 0\), which corresponds to the correct answer for the problem, option (A).