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

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 [2004] (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) \(x^{2}+y^{2}-2 x+2 y-23=0\).

Step 1: Find the Intersection Point

To find the center of the circle, we need to solve the system of equations given by the line equations. This involves solving the equations \(2x + 3y + 1 = 0\) and \(3x - y - 4 = 0\) simultaneously. Start by solving the second equation for \(y\): \(y = 3x - 4\). Substitute this value into the first equation, resulting in \(2x + 3(3x - 4) + 1 = 0\). Simplifying gives \(2x + 9x - 12 + 1 = 0\), so \(11x = 11\), thus \(x = 1\). Substitute back into \(y = 3x - 4\) to find \(y = -1\). Therefore, the intersection point is \((1, -1)\), which is the center of the circle.

Step 2: Find the Radius of the Circle

The radius \(r\) of the circle is determined by the circumference formula \(2\pi r = 10\pi\). Solving for \(r\), we get \(r = 5\).

Step 3: Use the Standard Circle Equation

The general equation for a circle in standard form is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius. Substituting the center \((1, -1)\) and radius \(5\), we get \((x - 1)^2 + (y + 1)^2 = 25\).

Step 4: Expand and Simplify the Equation

Expand \((x - 1)^2 + (y + 1)^2 = 25\) to obtain the standard quadratic form \(x^2 - 2x + 1 + y^2 + 2y + 1 = 25\). Simplify to get: \(x^2 + y^2 - 2x + 2y + 2 = 25\), which can be written as \(x^2 + y^2 - 2x + 2y - 23 = 0\).

Step 5: Identify the Correct Answer

Comparing this with the given options, we identify that the equation \(x^2 + y^2 - 2x + 2y - 23 = 0\) corresponds to option (A).

Key Concepts

Line Intersection

Understanding the concept of line intersection is essential when dealing with problems involving geometric shapes and equations of lines. In this exercise, we need to find the point where two lines intersect because this point becomes the center of the circle defined by these lines.

To find the intersection point, equations of both lines are solved simultaneously. This involves substituting variables to find their common solution.

• For example, consider the equations given in the exercise: \(2x + 3y + 1 = 0\) and \(3x - y - 4 = 0\).
• We can express one variable in terms of another from one equation. Here, from the second equation, we find \(y = 3x - 4\).
• Substitute this into the first equation to find \(x\), and then use this \(x\) value to find \(y\).

The intersection point \((1, -1)\) obtained from this substitution is crucial as it represents the center of the circle.

Circle Radius

The radius of a circle is a fundamental concept that connects to various features of a circle such as its diameter, circumference, and area.

In this exercise, we use the circle's circumference to find the radius. The circumference formula is given by \(2\pi r\), where \(r\) is the radius. From the exercise, we know the circumference is \(10\pi\).

• Set the formula: \(2\pi r = 10\pi\).
• To find \(r\), simplify this equation which gives \(r = 5\).

The radius \(5\) is important because it lets us write the circle's equation in standard form using the circle's equation parameters.

Standard Form of Circle Equation

The standard form of the circle's equation is a convenient way to represent any circle on the Cartesian plane. This form makes it straightforward to identify the circle's center and radius.

The standard equation is expressed as \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius.

• In the problem, we have determined the center to be \((1, -1)\) and the radius \(r = 5\).
• Substitute these into the equation to get: \((x - 1)^2 + (y + 1)^2 = 25\).
• Expanding and simplifying this equation gives: \(x^2 + y^2 - 2x + 2y - 23 = 0\).

This expression is the specific standard form for the circle described in the problem, allowing us to identify the correct solution among the given options.