If a line, y=m x+c is a tangent to the circle,

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

$\left|\frac{c+3}{\sqrt{2}}\right|=1$

$\Rightarrow \quad c=-3 \pm \sqrt{2}$

$\Rightarrow \quad(c+3)^{2}=2$

$\Rightarrow \quad c^{2}+6 c+9=2$

$\therefore \quad c^{2}+6 c+7=0$