Problem 113 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The lines \(2 x-3 y=5\) and \(3 x-4 y=7\) are diameters of a circle having area as 154 sq units. Then the equation of the circle is (A) \(x^{2}+y^{2}+2 x-2 y=62\) (B) \(x^{2}+y^{2}+2 x-2 y=47\) (C) \(x^{2}+y^{2}-2 x+2 y=47\) (D) \(x^{2}+y^{2}-2 x+2 y=62\)

Step-by-Step Solution

Answer

The equation of the circle is option (C): \(x^2+y^2-2x+2y=47\).

Step 1: Find the center of the circle

The center of the circle is the intersection point of the two diameters, given by the equations \(2x - 3y = 5\) and \(3x - 4y = 7\). We find it by solving this system of equations. Using the method of substitution or elimination, solve for \(x\) and \(y\). \( x = 1 \) and \( y = -1 \). Thus, the center of the circle is \((1, -1)\).

Step 2: Determine the radius of the circle

Use the formula for the area of circle \(\pi r^2 = 154\). Solving for \(r\), we get \(r^2 = \frac{154}{\pi}\). Given \(\pi \approx 3.14\), \(r^2 \approx 49\). The radius of the circle is 7 units.

Step 3: Write the circle's equation

The general formula for a circle with center \((h, k)\) and radius \(r\) is \((x-h)^2 + (y-k)^2 = r^2\). Here, \((h, k) = (1, -1)\) and \(r^2 = 49\). Therefore, \((x-1)^2 + (y+1)^2 = 49\).

Step 4: Expand and simplify the circle's equation

Expand \((x-1)^2 + (y+1)^2 = 49\) to get: \[ (x^2 - 2x + 1) + (y^2 + 2y +1) = 49 \] \[ x^2 + y^2 - 2x + 2y + 2 = 49 \]. Reduce this to \[ x^2 + y^2 - 2x + 2y = 47 \].

Step 5: Find the correct option

Comparing the expanded equation \(x^2 + y^2 - 2x + 2y = 47\) with given options, we find option (C) is identical to the results obtained.

Key Concepts

System of Equations

To find the center of a circle where two lines act as its diameters, we must determine their intersection point. The given lines are represented by the equations \(2x - 3y = 5\) and \(3x - 4y = 7\). This setup forms what is known as a 'system of equations', meaning multiple equations that are solved together to find their common solution.There are several methods to solve a system of equations:

• Substitution: Solve one equation for a variable and substitute it into the other equation.
• Elimination: Add or subtract equations to eliminate one variable, making it simple to solve for the other.

In this case, using either method will lead you to the intersection point \((x, y) = (1, -1)\). Thus, we found that the center of the circle is at coordinates (1, -1), which is after initializing and simplifying by the system of equations. This is a crucial step in determining the circle's equation.

Circle Geometry

In circle geometry, understanding the key properties of a circle is essential. A circle is defined by its center and radius. Given that the lines provided in the exercise are diameters, the center is at their intersection.With the center of the circle established at \((1, -1)\), the radius is now vital. The radius can be calculated using the area of the circle formula. Provided the area \(\pi r^2 = 154\), solving for \(r^2\) gives \(49\) (approximate \(\pi\) as \(3.14\)). Thus, the radius \(r\) is 7 units long.The general equation for a circle's geometry is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center. Here, substituting \(h=1\), \(k=-1\), and \(r=7\) results in the equation \((x-1)^2 + (y+1)^2 = 49\). This is the groundwork for describing the circle using its geometrical characteristics.

Circle Area Formula

The circle area formula \(\pi r^2\) plays a crucial role in finding the specifics of a given circle. This formula allows us to seamlessly transition between understanding the area and determining the radius.In this problem, we start with an area of 154 square units. Setting up the equation \(\pi r^2 = 154\) and solving for \(r^2\) provides insight into the radius' value squared. Using \(\pi \approx 3.14\) simplifies calculations to \(r^2 \approx 49\). Taking the square root, we find \(r = 7\) units, confirming the circle's size as reflected in its geometry description.Accurate use of the circle area formula bridges gaps between abstract geometric concepts and tangible numeric solutions, allowing students to compute and verify the circle's characteristics effectively.