The equation to the line joining the centres of the circles belonging to the coaxial system of circles $4…
The equation to the line joining the centres of the circles belonging to the coaxial system of circles
$4 x^2+4 y^2-12 x+6 y-3+\lambda(x+2 y-6)=0$
is
$8 x-4 y-15=0$
$8 x-4 y+15=0$
$3 x-4 y-5=0$
$3 x-4 y+5=0$
Solution
Given coaxial system of circle is
$\begin{aligned} & 4 x^2+4 y^2-12 x+6 y-3+\lambda(x+2 y-6)=0 \\ & \text { or } x^2+y^2-3 x+\frac{3 y}{2}-\frac{3}{4}+\frac{\lambda}{4}(x+2 y-6)=0\end{aligned}$
Here, radical axis is $x+2 y-6=0$. i.e., line of centre is
$2 x-y+k=0$
Here, centre of circle is $\left(\frac{3}{2},-\frac{3}{4}\right)$ which lies on Eq. (i).
$\begin{array}{rlrl} & \therefore & 2\left(\frac{3}{2}\right)+\frac{3}{4}+k & =0 \\ \Rightarrow & \frac{12+3}{4}+k & =0 \\ \Rightarrow & k & =-\frac{15}{4}\end{array}$
$\therefore$ Required equation is
$\begin{aligned} 2 x-y-\frac{15}{4} & =0 \\ \Rightarrow \quad 8 x-4 y-15 & =0\end{aligned}$