Problem 78 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

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 (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

(d) The equation of the circle is \(x^2 + y^2 - 2x + 2y - 47 = 0\).

Step 1: Find the Center of the Circle

Since the given lines are diameters, they must intersect at the center of the circle. Therefore, we need to solve the two linear equations simultaneously to find their intersection point. Solve:\[\begin{align*}3x - 4y - 7 &= 0 \2x - 3y - 5 &= 0\end{align*}\]Using substitution or elimination, solve for \(x\) and \(y\). Multiply the first equation by 3 and the second by 4:\[\begin{align*}9x - 12y - 21 &= 0 \8x - 12y - 20 &= 0\end{align*}\]Subtract the second equation from the first:\[x - 1 = 0 \Rightarrow x = 1\]Substitute \(x = 1\) back into the second original equation:\[2(1) - 3y - 5 = 0 \Rightarrow 2 - 3y = 5 \Rightarrow -3y = 3 \Rightarrow y = -1\]Thus, the center of the circle is \((1, -1)\).

Step 2: Find the Radius of the Circle

The area of the circle is given as \(49\pi\) square units. Recall the formula for the area of a circle: \[A = \pi r^2\]Equating the given area to the formula:\[49\pi = \pi r^2\]Cancel \(\pi\) from both sides:\[r^2 = 49\]Taking the square root of both sides:\[r = 7\]Therefore, the radius of the circle is 7 units.

Step 3: Write the Equation of the Circle

The standard form of the equation of a circle is \[(x - h)^2 + (y - k)^2 = r^2\]where \((h, k)\) is the center and \(r\) is the radius. Substituting the center \((1, -1)\) and radius 7 into the formula:\[(x - 1)^2 + (y + 1)^2 = 49\]Expand:\[x^2 - 2x + 1 + y^2 + 2y + 1 = 49\]Combine like terms:\[x^2 + y^2 - 2x + 2y + 2 = 49\]Simplify the equation:\[x^2 + y^2 - 2x + 2y - 47 = 0\]

Step 4: Compare and Choose the Correct Option

Now, compare the derived equation \(x^2 + y^2 - 2x + 2y - 47 = 0\) with the given options to find the correct match:(a) \(x^{2}+y^{2}+2x-2y-47=0\) (b) \(x^{2}+y^{2}+2x-2y-62=0\) (c) \(x^{2}+y^{2}-2x+2y-62=0\) (d) \(x^{2}+y^{2}-2x+2y-47=0\) The derived equation matches option (d). Therefore, the correct equation of the circle is \(x^{2}+y^{2}-2x+2y-47=0\).

Key Concepts

Intersection of Lines

When two lines intersect, they meet at a specific point in the coordinate plane. This point of intersection can be easily found by solving the equations of the two lines simultaneously.
In our exercise, the lines are given by the equations \(3x - 4y - 7 = 0\) and \(2x - 3y - 5 = 0\).
To find the intersection point, follow these steps:

• Use substitution or elimination methods to solve the system of equations.
• Multiply the first equation by 3 and the second by 4 to align coefficients of \(y\).
• Subtract one equation from the other to eliminate \(y\) and solve for \(x\).
• Substitute the value of \(x\) back into one of the original equations to find \(y\).

In this case, following these steps led us to the intersection at \((1, -1)\), which is crucial as it serves as the center of the circle where these lines are diameters.

Geometry in Coordinate Plane

Geometry in the coordinate plane deals with the placement and properties of shapes using a Cartesian system.
This exercise involves a circle, where the lines are diameters. This means the lines go through the center of the circle, and intersect each other at right angles.
The coordinate plane helps locate not just intersections but also visualize the layout of the circle concerning these lines. This makes calculating the center easier by solving intersections, as done previously.
Understanding this layout is essential. It helps not just in locating key points like the center, but also confirms that our geometric reasoning aligns with algebraic solutions.

Circle Area Formula

The circle area formula \(A = \pi r^2\) is fundamental in geometry. Here, it helps in determining the radius of the circle from its given area.
Our problem states an area of \(49\pi\) square units. Using the formula:

• Set the area \(49\pi = \pi r^2\).
• Cancel \(\pi\) from both sides, simplifying to \(r^2 = 49\).
• Take the square root of 49 to find \(r = 7\).

By using the area, we not only find the radius but also reinforce the properties of the circle related to its dimensions. Remember, the radius is key in writing the circle's equation.

System of Linear Equations

A system of linear equations consists of two or more equations that represent lines in the coordinate plane. Solving such a system provides points where the lines meet or intersect.
In problems involving circles and lines, like this one, solving these systems often helps find crucial circle properties like the center.
To solve:

• Use methods like elimination or substitution, aligning equations to simplify combining them.
• Find values of variables that satisfy all equations in the system.

By deriving the point \((1, -1)\), we confirmed the center of the circle, emphasizing the solution's reliance on understanding and manipulating these systems efficiently. Always ensure equations are balanced and solved methodically to find correct intersections.