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If the circle $x^2+y^2-6 x-12 y+1=0$ cuts another circle $\mathrm{C}$ orthogonally and the centre of the circle $\mathrm{C}$ is $(-4,2)$, then its radius is
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Verified Answer
The correct answer is:
$\sqrt{21}$
Given circles $x^2+y^2-6 x-12 y+1=0$ and $\mathrm{S}_1=0$ with centre $(-4,2)$ cut each other orthogonally
$$
\begin{aligned}
& \Rightarrow g_1=3, f_1=6, C_1=1 \\
& g_2=-4 f_2=2, C_2=x
\end{aligned}
$$
Since for two circles
$$
\begin{aligned}
& 2\left(\mathrm{~g}_1 \mathrm{~g}_2+\mathrm{f}_1 \mathrm{f}_2\right)=\mathrm{C}_1+\mathrm{C}_2 \\
& \Rightarrow \mathrm{C}_2=\mathrm{x}=-1
\end{aligned}
$$
Now $r=\sqrt{g_2^2+f^2-x} \Rightarrow \sqrt{16+4+1}=\sqrt{21}$
$$
\begin{aligned}
& \Rightarrow g_1=3, f_1=6, C_1=1 \\
& g_2=-4 f_2=2, C_2=x
\end{aligned}
$$
Since for two circles
$$
\begin{aligned}
& 2\left(\mathrm{~g}_1 \mathrm{~g}_2+\mathrm{f}_1 \mathrm{f}_2\right)=\mathrm{C}_1+\mathrm{C}_2 \\
& \Rightarrow \mathrm{C}_2=\mathrm{x}=-1
\end{aligned}
$$
Now $r=\sqrt{g_2^2+f^2-x} \Rightarrow \sqrt{16+4+1}=\sqrt{21}$
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