Question

A circle touches both the coordinate axes and the straight line \( L \equiv 4x + 3y - 6 = 0 \) in the first quadrant. If this circle lies below the line \( L = 0 \), then the equation of that circle is

• A. \( 4x^2 + 4y^2 - 4x - 4y + 1 = 0 \)
• B. \( 4x^2 + 4y^2 - 4x - 24y + 1 = 0 \)
• C. \( x^2 + y^2 - 6x - 6y + 9 = 0 \)
• D. \( x^2 + y^2 - 6x - y - 9 = 0 \)

Hint:

When a circle touches both coordinate axes in the first quadrant, its center is at \( (r, r) \). Use perpendicular distance from this point to the line to relate the radius and ensure tangency.

Answer 1

The Correct Option is A

Step 1: Circle touching coordinate axes.
If a circle touches both coordinate axes and lies in the first quadrant, its center is at \( (a, a) \), and radius is \( a \). So its equation is: \[ (x - a)^2 + (y - a)^2 = a^2 \] Step 2: Circle touches the line.
Given it also touches the line \( 4x + 3y - 6 = 0 \), the perpendicular distance from the center \( (a, a) \) to this line must be equal to the radius \( a \). Use the distance formula: \[ \frac{|4a + 3a - 6|}{\sqrt{4^2 + 3^2}} = a \Rightarrow \frac{|7a - 6|}{5} = a \] Solving: \[ |7a - 6| = 5a \Rightarrow \text{Two cases:} \] Case 1: \( 7a - 6 = 5a \Rightarrow 2a = 6 \Rightarrow a = 3 \) Case 2: \( -(7a - 6) = 5a \Rightarrow -7a + 6 = 5a \Rightarrow 12a = 6 \Rightarrow a = \frac{1}{2} \) Since the circle lies below the line \( L = 0 \), we choose the smaller \( a = \frac{1}{2} \) Step 3: Write the equation.
Substitute \( a = \frac{1}{2} \): \[ (x - \frac{1}{2})^2 + (y - \frac{1}{2})^2 = \left(\frac{1}{2}\right)^2 \Rightarrow x^2 + y^2 - x - y + \frac{1}{4} = \frac{1}{4} \Rightarrow x^2 + y^2 - x - y = 0 \] Multiply entire equation by 4 to match one of the options: \[ 4x^2 + 4y^2 - 4x - 4y = 0 + 0 \Rightarrow \boxed{4x^2 + 4y^2 - 4x - 4y + 1 = 0} \]