If the lines `a_1x+b_1y+c_1=0` and `a_2x+b_2y+c_2=0` cut the coordinae axes at concyclic points, then prove that `|a_1a_2|=|b_1b_2|`

Options

A. `(a_(1)x+b_(1)y+c_(1))(a_(2)x+b_(2)y+c_(2))+xy=0`
B. `(a_(1)x+b_(1)y+c_(1))(a_(2)x+b_(2)y+c_(2))+(a_(1)b_(2)+a_(2)b_(1))xy=0`
C. `(a_(1)x+b_(1)y+c_(1))(a_(2)x+b_(2)y+c_(2))-(a_(1)b_(2)+a_(2)b_(1))xy=0`
D. `(a_(1)x+b_(1)y+c_(1))(a_(2)x+b_(2)y+c_(2))-(a_(1)b_(2)-a_(2)b_(1))=0`

Answer

C

Explanation

The equation of the second degree curve passing through the points of intersection of the given lines with the coordinate axes is
`(a_(1)x + b_(1) y + c_(1)) (a_(2)x+b_(2)y+c_(2)) + lambda xy = 0 ...(i)`
This will represent a circle, if
Coefficient of `x^(2)=` Coefficient of `y^(2)` and, Coeff. of xy=0
`rArr a_(1)a_(2)=b_(1)b_(2) and a_(1)b_(2) + a_(2)b_(1) + lambda = 0`
`rArr a_(1)a_(2) = b_(1)b_(2) and lambda = - (a_(1)b_(2)+a_(2)b_(1)) xy=0`
as the equation of the required circle.