Question

The centre of family of circles cutting the family of circles $x^2+y^2+4x(\lambda -\frac{3}{2})+3y(\lambda -\frac{4}{3})-6(\lambda +2)=0$ orthogonally, lies on

• A. $x-y-1=0$
• B. $4x+3y-6=0$
• C. $4x+3y+7=0$
• D. $3x-4y-1=0$

Answer 1

The correct option is D $3x-4y-1=0$

$S+\lambda L=0⟹L=4x+3y-6=0;S=x^2+y^2-6x-4y-12=0$

$⟹(3,2)$ is the centre and the other centre should be on the line with this centre such that it is perpendicular to $L$

$y-2=\frac{3}{4}(x-3)⟹3x-4y-1=0$