$x^2+y^2-6x-6y+4=0$ $x^2+y^2-2x-4y+3=0$ $x^2+y^2+2kx+2y+1=0$ If the radical centre of the above three circles exists, then which of the following cannot be the value of  $k$ ?

$\begin{array}{r}{\mathrm{S}}_1:x^2+y^2-6x-6y+4=0\\ {\mathrm{S}}_2:x^2+y^2-2x-4y+3=0\\ {\mathrm{S}}_3:x^2+y^2+2kx+2y+1=0\end{array}$

Now radical centre of S1 and S2 is given by 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{S}_1-S_2=0\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4x-2y+1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4x+2y-1=0\end{array}$

Radical center of S2 and S3 is given by 

$\begin{array}{l}{S}_2-S_3=0\\ -2(1+k)x-6y+2=0\\ (1+k)x+3y-1=0\end{array}$

$k=1:2x+3y-1=0$

$\begin{array}{r}k=2:3x+3y-1=0\\ k=4:5x+3y-1=0\\ k=5:6x+3y-1=0\end{array}$

Only $6x+3y-1=0$ is parallel to $4x+2y-1=0$

So for k=5 radical center does not exists

$\therefore k\ne 5$