The equation to the line joining the centres of the circles belonging to the coaxial system of circles
` 4x^(2) + 4y^(2) - 12x + 6y - 3 + lambda (x + 2y - 6) = 0 ` is

Options

A. 8x -4y - 15 = 0
B. 8x - 4y + 15 = 0
C. 3x - 4y - 5 = 0
D. 3x - 4y + 5 = 0

Answer

A

Explanation

Given coaxial system of circle is
`4x^(2) + 4y^(2) - 12x + 6y - 3 + lambda (x + 2y - 6) = 0`
`implies` centre of S = 0 is `C ((3)/(2), (-3)/(4))`
`implies` Equation of line perpendicular to
x + 2y - 6 is, 2x - y + k = 0
Equation (1) passes through C `((3)/(2), (-3)/(4))`
`implies 3 + (3)/(4) + k = 0`
`implies k = (-15)/(4)`
`implies` Equation (1) becomes, `2x - y -(15)/(4) = 0`
`:. 8x - 4y - 15 = 0`