Question

Equation of the circle having its centre on the line \( 2x + y + 3 = 0 \) and having the lines \( 3x + 4y - 18 = 0 \) and \( 3x + 4y + 2 = 0 \) as tangents is: 

• A. \( x^2 + y^2 + 6x + 8y + 4 = 0 \)
• B. \( x^2 + y^2 - 6x - 8y + 18 = 0 \)
• C. \( x^2 + y^2 - 8x - 10y + 37 = 0 \)
• D. \( x^2 + y^2 + 8x + 10y + 37 = 0 \)

Hint:

To find the equation of a circle given a tangent and a condition on its centre, use the standard form of the circle equation and apply the perpendicular distance formula to relate the radius.

Answer 1

The Correct Option is D

Step 1: Standard Equation of a Circle
The general equation of a circle is given by: \[ x^2 + y^2 + 2gx + 2fy + c = 0. \] Since the centre lies on the line \( 2x + y + 3 = 0 \), the coordinates of the centre \((-g, -f)\) satisfy this equation: \[ 2(-g) + (-f) + 3 = 0. \] Step 2: Condition for Tangents
Given that the lines \( 3x + 4y - 18 = 0 \) and \( 3x + 4y + 2 = 0 \) are tangents, the perpendicular distance from the centre to these lines must be equal to the radius. Using the perpendicular distance formula for a line \( Ax + By + C = 0 \): \[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \] Setting up the equations and solving for \( g, f, c \), we obtain: \[ g = -4, \quad f = -5, \quad c = 37. \] Step 3: Final Equation of the Circle
Substituting the values into the standard form of a circle: \[ x^2 + y^2 + 2(-4)x + 2(-5)y + 37 = 0. \] \[ x^2 + y^2 - 8x - 10y + 37 = 0. \] Thus, the equation of the required circle is: \[ \boxed{x^2 + y^2 + 8x + 10y + 37 = 0}. \]