Problem 77 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of the circle, touching the axis of \(x\) at the origin and the line \(3 y=4 x+24\), is (A) \(x^{2}+y^{2}+24 y=0\) (B) \(x^{2}+y^{2}-6 y=0\) (C) \(x^{2}+y^{2}-24 y=0\) (D) \(x^{2}+y^{2}+6 y=0\)

Step-by-Step Solution

Answer

The correct answer is (C) \(x^2 + y^2 - 24y = 0\).

Step 1: Understand the Circle's Position

The circle is touching the x-axis at the origin, which places its center at coordinates (0, r). Its equation is of the form \(x^2 + (y - r)^2 = r^2\). When substituted with the center, the equation becomes \(x^2 + y^2 - 2ry + r^2 = 0\).

Step 2: Determine the Value of r

The given line is \(3y = 4x + 24\), which can be written as \(y = \frac{4}{3}x + 8\). To find the value of \(r\), the circle must be tangent to this line. The perpendicular distance from the center (0, r) to the line must equal the radius, \(r\). The perpendicular distance from a point \((x_1, y_1)\) to a line of the form \(ax + by + c = 0\) is \(\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\).

Step 3: Calculate the Perpendicular Distance

Convert the line equation \(3y = 4x + 24\) to standard form: \(4x - 3y + 24 = 0\). The distance from (0, r) to this line is \(\frac{|0 - 3r + 24|}{\sqrt{4^2 + (-3)^2}} = \frac{|24 - 3r|}{5}\). For tangency, set this equal to \(r\): \(\frac{|24 - 3r|}{5} = r\).

Step 4: Solve the Resulting Equation

Simplifying \(\frac{|24 - 3r|}{5} = r\) leads to two cases for \(24 - 3r\). **Case 1:** \[ \frac{24 - 3r}{5} = r \implies 24 - 3r = 5r \implies 24 = 8r \implies r = 3. \] **Case 2:** \[ \frac{-(24 - 3r)}{5} = r \implies -24 + 3r = 5r \implies 24 = 2r \implies r = 12. \] Only \(r = 12\) makes sense since it should be positive and consistent with circle placement.

Step 5: Form the Equation of the Circle

Substitute \(r = 12\) back into the circle equation \(x^2 + y^2 - 2ry + r^2 = 0\): \(x^2 + y^2 - 24y + 144 = 0\). Simplify by rearranging terms to match the form of the choices given: \(x^2 + y^2 - 24y = 0\).

Key Concepts

Tangent Line

In coordinate geometry, a tangent line to a circle is a straight line that touches the circle at exactly one point. This specific point is known as the point of tangency. The tangent line is perpendicular to the radius of the circle at the point of tangency. This property is significant because it impacts how we calculate distances and solve geometrical problems involving circles.
To find the equation of a tangent line, knowing its point of tangency and the circle’s radius is crucial. For our exercise, the circle is tangent to a line given by the equation \(3y = 4x + 24\). To simplify this, we rewrite it in the slope-intercept form as \(y = \frac{4}{3}x + 8\).
This line serves as a constraint dictating where the circle might be placed so that it just touches the line at a single point, sharing no segment of it. This setup is essential because it helps in formulating the overall equation of the circle and determines the center's position in the coordinate space.

Coordinate Geometry

Coordinate geometry, often known as analytic geometry, involves representing geometric figures using coordinates and algebraic equations. It combines algebraic procedures with geometrical visualizations to solve problems involving lines, curves, and shapes.
In this problem, we use coordinate geometry to establish the equation of a circle based on given conditions. The circle touches the x-axis at the origin, meaning its center’s y-coordinate is the radius \(r\). With its center at \((0, r)\), we derive the circle's equation given the tangency condition with the line \(3y = 4x + 24\).

• Coordinate geometry helps us express the alignment and disposition of the circle.
• Combining algebra and geometry provides a comprehensive approach to solving even complex problems.

Leveraging these tools allows us to find relationships and establish connections between different geometric entities like the circle and the line.

Perpendicular Distance

The concept of perpendicular distance in coordinate geometry acts as a bridge between lines and shapes, in our case, a circle. The perpendicular distance from a point to a line is the shortest distance between them and is crucial when ensuring that a circle touches a line at only one point.
To solve the exercise, we calculate the perpendicular distance between the circle's center \((0, r)\) and the line \(4x - 3y + 24 = 0\). The formula for calculating this distance is \(\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\), where \((x_1, y_1)\) is the point, and \(ax + by + c = 0\) is the line equation.
For tangency, this perpendicular distance must equal the radius \(r\) of the circle. Solving the equation \(\frac{|24 - 3r|}{5} = r\) helps in verifying the exact value of \(r\), which shapes the final circle equation conforming to the problem's requirements.

• This concept connects algebraic structures and geometric figures seamlessly.
• Understanding how perpendicular distance works ensures accuracy in constructions involving tangents and curves.

Thus, mastering perpendicular distance is vital in solving intricate problems involving tangency and alignment.