If $L_1$ represents the radical axis of circles $x^2+y^2-4 x-6 y+5=0$ and $x^2+y^2-2 x-4 y-1=0$ and $L_2$…
If $L_1$ represents the radical axis of circles $x^2+y^2-4 x-6 y+5=0$ and $x^2+y^2-2 x-4 y-1=0$ and $L_2$ represents the radical axis of $x^2+y^2+2 x+2 y-7=0$ and $x^2+y^2+x+y+9=0$, then
$L_1$ is parallel to $L_2$.
$L_1$ is perpendicular to $L_2$.
$L_1$ and $L_2$ intersect at an angle $30^{\circ}$.
$L_1$ and $L_2$ intersect at $(1,7)$.
Solution
Circles $\rightarrow x^2+y^2-4 x-6 y+5=0 \quad \ldots\left(S_1\right)$
and $x^2+y^2-2 x-4 y-1=0$
Equation of radical axis is
$
\begin{aligned}
& S_1-S_2=0 \\
& L_1:\left(x^2+y^2-4 x-6 y+5\right)- \\
& \quad\left(x^2+y^2-2 x-4 y-1\right)=0
\end{aligned}
$
$
\begin{array}{ll}
& -2 x-2 y+6=0 \\
\Rightarrow & L_1: x+y-3=0 \\
\text { Similarly, } & L_2:\left(x^2+y^2+2 x+2 y-7\right)- \\
& \left(x^2+y^2+x+y+9\right)=0 \\
\Rightarrow & L_2: x+y-16=0
\end{array}
$
From Eqs. (i) and (ii), we obtain
$
\left\{m=\frac{\text { coefficient of } n}{\text { coefficient of } y}\right\}
$
$m_{L_1}=m_{L_2}=-1 \Rightarrow L_1$ and $L_2$ are parallel