If a line, $y=m x+c$ is a tangent to the circle, $(x-3)^{2}+y^{2}=1$ and it is perpendicular to a line $L_{1}$, where $L_{1}$ is the tangent to the circle, $x^{2}+y^{2}=1$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$; then:
Correct Option: , 3
Slope of tangent of $x^{2}+y^{2}=1$ at $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$\frac{1}{\sqrt{2}} x+\frac{1}{\sqrt{2}} y-1=0$
$x+y \sqrt{2}=0$, which is perpendicular to $x-y+c=0$
At $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ which is tangent of $(x-3)^{2}+y^{2}=1$
So, $m=1 \Rightarrow y=x+c$
Now, distance of $(3,0)$ from $y=x+c$ is
$\left|\frac{c+3}{\sqrt{2}}\right|=1$
$\Rightarrow \quad c=-3 \pm \sqrt{2}$
$\Rightarrow \quad(c+3)^{2}=2$
$\Rightarrow \quad c^{2}+6 c+9=2$
$\therefore \quad c^{2}+6 c+7=0$