Question

The line $x+2y+3=0$ cuts the circle $x^2+y^2+4x+4y-1=0$ at points $P$ and $Q$ and the line $2x+3y+\lambda =0$ cuts the circle $x^2+y^2+6x+2y-7=0$ at points $R$ and $S.$ If $P,$ $Q,$ $R$ and $S$ are concyclic, then the value of $\lambda$ is

Answer 1

Family of circles passing through the point of intersection of the circle $x^2+y^2+4x+4y-1=0$ and the line $x+2y+3=0,$ is
$(x^2+y^2+4x+4y-1)+t(x+2y+3)=0$
$\Rightarrow x^2+y^2+(4+t)x+(4+2t)y+(3t-1)=0\cdots \left(1\right)$

Similarly, family of circles through the other circle and line is
$(x^2+y^2+6x+2y-7)+s(2x+3y+\lambda )=0$
$\Rightarrow x^2+y^2+(6+2s)x+(2+3s)y+(\lambda s-7)=0...\left(2\right)$

As the intersection points are concylic, $\left(1\right)$ and $\left(2\right)$ represent the same circle.
So, $4+t=6+2s$ and $4+2t=2+3s$
On solving, we get $s=-6,t=-10$

Also, $\lambda s-7=3t-1$
$\Rightarrow \lambda (-6)-7=3(-10)-1$
$\Rightarrow 6\lambda =24$
$\Rightarrow \lambda =4$