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The radical centre of the circles $x^2+y^2-4 x-6 y+5=0$, $x^2+y^2-2 x-4 y-1=0$ and $x^2+y^2-6 x-2 y=0=0$ lies on the line
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Verified Answer
The correct answer is:
$18 x-12 y+1=0$
We have,
$$
\begin{aligned}
& S_1=x^2+y^2-4 x-6 y+5=0 \\
& S_2=x^2+y^2-2 x-4 y-1=0 \\
& S_3=x^2+y^2-6 x-2 y=0 \\
& S_1-S_2=0 \Rightarrow 2 x+2 y-6=0 \Rightarrow x+y-3=0 \\
& S_2-S_3=0 \Rightarrow 4 x-2 y-1=0
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
x=\frac{7}{6}, y=\frac{11}{6}
$$
$\left(\frac{7}{6}, \frac{11}{6}\right)$ satisfies the equations $18 x-12 y+1=0$
$$
\begin{aligned}
& S_1=x^2+y^2-4 x-6 y+5=0 \\
& S_2=x^2+y^2-2 x-4 y-1=0 \\
& S_3=x^2+y^2-6 x-2 y=0 \\
& S_1-S_2=0 \Rightarrow 2 x+2 y-6=0 \Rightarrow x+y-3=0 \\
& S_2-S_3=0 \Rightarrow 4 x-2 y-1=0
\end{aligned}
$$
Solving Eqs. (i) and (ii), we get
$$
x=\frac{7}{6}, y=\frac{11}{6}
$$
$\left(\frac{7}{6}, \frac{11}{6}\right)$ satisfies the equations $18 x-12 y+1=0$
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