Question

The line $x+2y+a=0$ intersects the circle $x^2+y^2-4=0$ at two distinct points $A$ and $B$. Another line $12x-6y-41=0$ intersects the circle $x^2+y^2-4x-2y+1=0$ at two distinct points $C$ and $D$.
The value of $a$ for which the four points $A,B,C$ and $D$ are concyclic is

Answer 1

$S+\lambda L=0$

$x^2+y^2-4+{\lambda }_1(x+2y+a)=0$

$x^2+y^2+{\lambda }_{1}x+2{\lambda }_{1}y-14-a\lambda 1)=0...(i)$

$x^2+y^2-4x-2y+1+{\lambda }_2(12x-6y-41)=0$

$\therefore x^2+y^2+(12{\lambda }_2-4)x+y(-6{\lambda }_2-2)-41{\lambda }_2+1=0...\left(ii\right)$

Equation the coefficient of (i) and (ii)

$\frac{1}{1}=\frac{1}{1}=\frac{{\lambda }_1}{12{\lambda }_2-4}=\frac{2\lambda 1}{-6{\lambda }_2-2}=\frac{-4+a{\lambda }_1}{1-41{\lambda }_2}$

$-6y_2=2=24{\lambda }_2-8$

$\Rightarrow 6=30{\lambda }_2$

$\therefore {\lambda }_2=\frac{1}{5}$

${\lambda }_1=12{\lambda }_2-4$

${\lambda }_1=\frac{12}{5}-4=\frac{-8}{5}$

$1-41{\lambda }_2=-4+a{\lambda }_1$

$\Rightarrow 1-\frac{41}{5}=-4-\frac{8a}{5}$

$\Rightarrow (5-\frac{41}{5})=\frac{8a}{5}$

$\Rightarrow +\frac{16}{5}=\frac{8a}{5}$

$\therefore a=+2$

$\therefore a=2$