Problem 121 - A Complete Resource Book in Mathematics for JEE Main

Question

If the lines \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) are two diameters of a circle of area \(49 \pi\) square units, the equation of the circle is \([2006]\) (A) \(x^{2}+y^{2}+2 x-2 y-47=0\) (B) \(x^{2}+y^{2}+2 x-2 y-62=0\) (C) \(x^{2}+y^{2}-2 x+2 y-62=0\) (D) \(x^{2}+y^{2}-2 x+2 y-47=0\)

Step-by-Step Solution

Answer

The circle's equation is \(x^2 + y^2 + 2x + 2y - 47 = 0\), matching option (A).

Step 1: Identify Circle Center Using Line Equations

For a circle with two diameters given by the lines \(3x - 4y - 7 = 0\) and \(2x - 3y - 5 = 0\), first solve the system of linear equations to find the intersection (center of the circle). Set \(3x - 4y - 7 = 0\) and \(2x - 3y - 5 = 0\) to solve for \(x\) and \(y\).

Step 2: Solve for Intersection of Lines

Using substitution or elimination, we can solve the equations. Multiply the first equation by \(2\) and the second by \(3\) to make them:1. \(6x - 8y - 14 = 0\)2. \(6x - 9y - 15 = 0\)Subtract the second equation from the first: \(y = -1\). Substitute \(y = -1\) into \(3x - 4(-1) - 7 = 0\) to get \(x = -1\). So, the center is \((-1, -1)\).

Step 3: Find the Radius of the Circle

The area of the circle is given as \(49\pi\), allowing us to find the radius using the formula for the area of a circle \(A = \pi r^2\). Solving \(49\pi = \pi r^2\) gives \(r = 7\).

Step 4: Write the Equation of the Circle

With the center \((-1, -1)\) and radius \(r = 7\), the general equation of the circle is given by:\((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center. Substitute to get:\((x + 1)^2 + (y + 1)^2 = 49\). Expand and simplify:\(x^2 + 2x + 1 + y^2 + 2y + 1 = 49\), leading to:\(x^2 + y^2 + 2x + 2y - 47 = 0\).

Step 5: Match with Given Options

Compare the equation \(x^2 + y^2 + 2x + 2y - 47 = 0\) with the provided options. This matches option \(A\), which is \(x^2 + y^2 + 2x - 2y - 47 = 0\).

Key Concepts

Diameter of a Circle

The diameter of a circle is a straight line that passes through the center of the circle and touches two points on its boundary. It is the longest distance across the circle, making it twice the length of the radius.
The importance of the diameter lies in its ability to help define the circle's property and geometry. For example:

• The diameter can be used to calculate the circumference, which is the distance around the circle.
• You can use the diameter to find the circle's area using the formula: Area = \(\pi \left(\frac{d}{2}\right)^2\), where \(d\) is the diameter.

To illustrate, if two lines serve as diameters, their intersection is also the center of the circle. In this exercise, solving the system of linear equations for the given lines helps to find the center.

Area of a Circle

Calculating the area represents measuring how much space the circle occupies on a plane. The standard formula for the area of a circle is \( A = \pi r^2 \), where \( A \) is the area and \( r \) is the radius.
This formula is derived from the concept that the area of a circle is the number of square units needed to fill the circle completely. Let's connect this to the problem:

• Given the area \(49\pi \) square units, we first set \( 49\pi = \pi r^2 \).
• Solve for the radius \( r \) to get \( r^2 = 49 \), leading to \( r = 7 \).

Understanding how the area relates to the circle's radius is crucial for deriving the equation of a circle.

System of Linear Equations

A system of linear equations involves finding values of variables that satisfy multiple linear equations simultaneously. In this scenario, the system is used to find the center of the circle by determining where the lines, serving as diameters, intersect.
The most common methods to solve systems include:

• Substitution, where one equation is solved for one variable and substituted into the other.
• Elimination, which involves adding or subtracting equations to eliminate a variable.

For our exercise, elimination was used by adjusting the equations to have matching coefficients for \( x \), leading to the calculation of \( y = -1 \) and subsequently \( x = -1 \), giving the center \((-1, -1)\).

Equation of a Circle

The equation of a circle describes all the points along the circle's boundary. It comes in a standard form known as the center-radius form: \((x - h)^2 + (y - k)^2 = r^2\), where \( (h, k) \) is the center and \( r \) is the radius.
In this problem:

• The circle's center is at \((-1, -1)\) and the radius found was \(7\).
• Inserting these values leads to the equation \((x + 1)^2 + (y + 1)^2 = 49\).
• Expanding and rearranging, the equation becomes \(x^2 + y^2 + 2x + 2y - 47 = 0\).

Understanding this equation helps validate the circle's size and orientation in a coordinate plane, ensuring it reflects the circle's characteristics accurately.