If a line, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is a tangent to the circle, $(\mathrm{x}-3{)}^2+{\mathrm{y}}^2=1$ and it is perpendicular to a line L₁ , where $L_1$ is the tangent to the circle, ${\mathrm{x}}^2+{\mathrm{y}}^2=1$ at the point $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);$ then:

Equation of tangent  at $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\text{ of }{\mathrm{x}}^2+{\mathrm{y}}^2=1 {\text{is S}}_{\text{1}}\text{=0}$

$\frac{1}{\sqrt{2}}\mathrm{x}+\frac{1}{\sqrt{2}}\mathrm{y}-1=0\Rightarrow \mathrm{x}+\mathrm{y}-\sqrt{2}=0,$      

$y=mx+c$ is perpendicular to $y=-x+\sqrt{2}$

Hence m=1

Now $y=mx+c$ is tangent of $(\mathrm{x}-3{)}^2+{\mathrm{y}}^2=1$,

whose centre=$\left(3,0\right)$  radius=1

     r=d

Now, distance of (3,0) from $\mathrm{y}=\mathrm{x}+\mathrm{c}\text{ is }$$\left|\frac{\mathrm{c}+3}{\sqrt{2}}\right|=1$

$\begin{array}{l}\Rightarrow \mathrm{c}=-3\pm \sqrt{2}\Rightarrow (\mathrm{c}+3{)}^2=2\Rightarrow {\mathrm{c}}^2+6\mathrm{c}+9=2\\ \therefore {\mathrm{c}}^2+6\mathrm{c}+7=0\end{array}$