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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
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Verified Answer
The correct answer is:
$L_1$ is parallel to $L_2$.
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.
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.
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