\(x^{2}+y^{2}-6 x-6 y+4=0, \quad x^{2}+y^{2}-2 x\) \(-4 y+3=0, x^{2}+y^{2}+2 k x+2 y+1=0\). If the radical centre of the above three circles exists, then which of the following cannot be the value of \(k\) ?

Given equation of circles are
\(S_{1} \equiv x^{2}+y^{2}-6 x-6 y+4=0\) \(\quad S_{2} \equiv x^{2}+y^{2}-2 x-4 y+3=0\) and \(S_{3} \equiv x^{2}+y^{2}+2 k x+2 y+1=0\) Now, radical axis of circle \(S_{1}\) and \(S_{2}\) is \[ S_{1}-S_{2}=0 \] \(\Rightarrow x^{2}+y^{2}-6 x-6 y+4-x^{2}-y^{2}+2 x\) \(\Rightarrow \quad-4 x-2 y+1=0\) \(\Rightarrow \quad 4 x+2 y-1=0\) Radical axis of circle \(S_{2}\) and \(S_{3}\) is \(-3=0\) \(\quad S_{2}-S_{3}=0\) \(\Rightarrow x^{2}+y^{2}-2 x-4 y+3-x^{2}-y^{2}-2 k x\) \(\Rightarrow \quad-(2+2 k) x-6 y+2=0\) \(\Rightarrow \quad(2+2 k) x+6 y-2=0\)
\[
\begin{array}{lr}
\Rightarrow & 24-4-4 k \neq 0 \\
\Rightarrow & 20-4 k \neq 0 \\
\Rightarrow & k \neq 5
\end{array}
\]