Problem 87 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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) \(x^{2}+y^{2}+2 x-2 y=62\) (c) \(x^{2}+y^{2}+2 x-2 y=47\) (d) \(x^{2}+y^{2}-2 x+2 y=47\).

Step-by-Step Solution

Answer

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

Step 1: Find the Coordinates of the Center

Since the lines are diameters of the circle, their point of intersection will be the center. Solve the equations:1. \(2x - 3y = 5\)2. \(3x - 4y = 7\).Multiply the first equation by 3:\(6x - 9y = 15\).Multiply the second equation by 2:\(6x - 8y = 14\).Subtract the second from the first:\(-9y + 8y = 15 - 14\) yields \(y = -1\).Substitute \(y = -1\) back into the first equation:\(2x - 3(-1) = 5\)\(2x + 3 = 5\)\(2x = 2\)\(x = 1\).Thus, the center of the circle is \((1, -1)\).

Step 2: Calculate the Radius using the Area

The area of the circle is given as 154 square units. The formula for the area of a circle is \(\pi r^2\), where \(r\) is the radius. Set up the equation:\[\pi r^2 = 154\]Solve for \(r^2\):\[r^2 = \frac{154}{\pi}\]\(\pi\) is approximately 3.14, so:\[r^2 = \frac{154}{3.14} \approx 49\].Thus, the radius squared \(r^2\) is 49.

Step 3: Write the Equation of the Circle

The general equation for a circle with center \((h, k)\) and radius squared \(r^2\) is:\[(x - h)^2 + (y - k)^2 = r^2\]Here, \(h = 1\) and \(k = -1\), and we found \(r^2 = 49\) from Step 2. Substitute these into the equation:\[(x - 1)^2 + (y + 1)^2 = 49\]Expand:\[(x^2 - 2x + 1) + (y^2 + 2y + 1) = 49\]Simplify:\[x^2 + y^2 - 2x + 2y + 2 = 49\]\[x^2 + y^2 - 2x + 2y = 47\].

Key Concepts

Coordinate Geometry

Coordinate geometry is a sector of mathematics that involves graphing and calculating the spatial relationships of geometric figures using an XY-plane. When solving problems related to the equation of a circle, coordinate geometry provides valuable tools. To identify the precise location of a point, it uses numerical coordinates, usually shown as \(x, y\).

In this specific exercise, the lines defined by the equations \(2x - 3y = 5\) and \(3x - 4y = 7\) serve as the diameters of a circle. To find the center of the circle, one must find the intersection of these two lines. By solving these equations simultaneously using methods such as substitution or elimination, we can find their intersection at the point (1, -1).

This intersection point is significant because it represents the center of the circle. In coordinate geometry, this point becomes a key to forming the equation of the circle, guiding us in placing the circle accurately within the XY-plane.

Geometry Problems

Geometry problems often involve finding unknown values or measurements using specific geometric properties or rules. In this problem, the intersection of the diameters serves as the center of the circle. This solution involves recognizing this key property that diameters of a circle intersect at its center, helping to resolve the problem quickly.

After identifying the center, the next step is to utilize the given area of the circle, which is crucial for determining the circle's size or scale. The problem implies that knowledge of geometry allows for connecting the provided area to the calculation of the radius.

Solving this geometry problem involves using several geometric principles:

• Understanding that the intersection point of the lines represents the circle center.
• Recognizing the relationship between the area and radius (via \(\pi r^2 = ext{Area}\)).
• Confirming the value of the radius squared as 49, significantly affecting the final circle equation.

Using these principles, various problem components come together harmoniously to generate a coherent solution.

Circle Area Calculation

The calculation of a circle's area is fundamental in understanding circular shapes in geometry. The equation for the area \(\pi r^2\) takes the radius as input, reflecting how altering the radius affects the area. Being given an area allows us to work backward to find the radius.

In this exercise, you start with a circle area of 154 square units. Using the formula \(\pi r^2 = 154\), you solve for \(r^2\) by dividing the area by \(\pi\). Approximating \(\pi\) to 3.14 simplifies the calculation, yielding a radius squared \(r^2 = 49\) based on \(\frac{154}{3.14}\).

Knowing \(r^2\) is vital for forming the circle’s equation. Together with the center coordinates, it allows you to express the circle’s geometry through its formula. This problem illustrates how understanding the area and radius relationship leads to constructing accurate circle equations. It's not only about knowing the formula, but understanding how each part affects the whole.