Problem 61 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the circle \(x^{2}+y^{2}-6 x-8 y+\left(25-a^{2}\right)=0\) touches the axis of \(x\), then a equals. (a) 0 (b) \(\pm 4\) (c) \(\pm 2\) (d) \(\pm 3\)

Step-by-Step Solution

Answer

The value of \(a\) is \(\pm 4\).

Step 1: Identify the General Equation of a Circle

The given equation of the circle is \(x^{2} + y^{2} - 6x - 8y + (25 - a^{2}) = 0\). The general form of a circle's equation is \((x-h)^{2} + (y-k)^{2} = r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.

Step 2: Complete the Square

To express the circle's equation in standard form, complete the square for \(x\) and \(y\). First, for \(x\), the terms are \(x^{2} - 6x\). Half of \(-6\) is \(-3\), so add and subtract \(9\). For \(y\), the terms are \(y^{2} - 8y\). Half of \(-8\) is \(-4\), so add and subtract \(16\).

Step 3: Rewrite the Completed Square Equation

The equation becomes \((x-3)^{2} + (y-4)^{2} = a^{2}\). The constant terms \(-9\) and \(-16\) are moved to the other side with \(25 - a^{2}\), leading to \((x-3)^{2} + (y-4)^{2} = 25 - a^{2} + 25\).

Step 4: Determine the Circle's Properties

From the standard form, the center of the circle is \((3, 4)\) and the radius squared is \(a^{2}\). The equation becomes \((x-3)^{2} + (y-4)^{2} = a^{2}\).

Step 5: Condition for a Circle to Touch the x-axis

For the circle to touch the x-axis, the distance from the center to the x-axis must equal the radius. The distance from the center \((3, 4)\) to the x-axis is \(4\). Therefore, the radius is \(4\), so \(a^{2} = 16\).

Step 6: Solve for a

Given that \(a^{2} = 16\), we find \(a = \pm 4\). Therefore, the possible values for \(a\) that satisfy the condition are \(4\) or \(-4\).

Key Concepts

Coordinate Geometry

Coordinate geometry is a branch of mathematics that allows us to study shapes and figures using coordinates and algebra. This powerful tool combines geometric and algebraic techniques to solve problems involving points, lines, and shapes on the Cartesian plane.

In coordinate geometry, we use coordinates to represent points. Each point on the plane has an ordered pair \(x, y\), where \(x\) is the horizontal coordinate, and \(y\) is the vertical one.

• Coordinates help to describe positions and determine distances between points.
• Equations in coordinate geometry help in defining and analyzing various geometric figures like lines, circles, and parabolas.

For example, the equation \(x^{2} + y^{2} - 6x - 8y + (25 - a^{2}) = 0\) represents a circle in the coordinate plane, which can be analyzed further using coordinate geometry techniques.

Equation of a Circle

The equation of a circle is a mathematical expression that describes all the points equidistant from a fixed point, called the center. In its standard form, the equation is \((x-h)^2 + (y-k)^2 = r^2\) where \(h, k\) are the coordinates of the center, and \(r\) is the radius.

This equation allows us to quickly determine key properties of the circle, such as:

• The center at point \( (h, k) \)
• The radius, given by \( r \)

By converting the equation \(x^{2} + y^{2} - 6x - 8y + (25 - a^{2}) = 0\) into its standard form, we identify the center as \( (3, 4) \), and the radius as \( a\).

Each term in the circle's equation helps to pinpoint a specific characteristic, such as shift from the origin which is critical in contextual geometric problems.

Completing the Square

Completing the square is a method used in algebra to transform a quadratic equation into a perfect square trinomial. This process is essential for rewriting circle equations into the standard form \((x-h)^2 + (y-k)^2 = r^2\).

Here's a quick step-by-step summary:

• Start with the terms \(x^2 + bx\) from the equation.
• Find half of \(b\) and square it, then add and subtract this square inside the equation.
• Repeat the same steps for \(y^2 + dy\).

In our equation, we completed the square twice:

• For \(x\): \(x^2 - 6x\) turned into \((x-3)^2\) by adding \((\frac{-6}{2})^2 = 9\).
• For \(y\): \(y^2 - 8y\) turned into \((y-4)^2\) by adding \((\frac{-8}{2})^2 = 16\).

By using this method, equations become more manageable, facilitating the understanding of the circle's attributes.

Axis Tangency Condition

The tangency condition determines when a curve like a circle touches a specific line, such as one of the coordinate axes, without crossing it. This condition is crucial for solving geometric problems where precision in positioning on the axes is required.

For a circle to touch the x-axis, the distance from the center of the circle to the x-axis must equal the radius of the circle.

Here's what this implies:

• The center's y-coordinate, which is the vertical distance from the center to the x-axis, must be equal to the length of the radius for tangency.

In our problem, with the circle centered at \( (3, 4) \), the y-coordinate is \(4\). Therefore, for tangency along the x-axis, the radius must also be \(4\).

• This implies \(a^2 = 16\) leading to \(a = ±4\), demonstrating the circle's perfect touch along the x-axis without overlapping it.