If a line $$y = mx + c$$, 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

We are told that a line $$y = mx + c$$ is tangent to the circle $$(x - 3)^2 + y^2 = 1$$ and that this line is perpendicular to another line $$L_1$$, which itself is the tangent to the circle $$x^2 + y^2 = 1$$ at the point $$\Bigl(\dfrac1{\sqrt2}, \dfrac1{\sqrt2}\Bigr)$$. Our task is to determine the relation satisfied by the constant $$c$$.

We begin with the tangent $$L_1$$ to the circle $$x^2 + y^2 = 1$$. A standard result for a circle of the form $$x^2 + y^2 = r^2$$ is that the tangent at any point $$(x_1,y_1)$$ on the circle is given by $$x x_1 + y y_1 = r^2$$. Here $$r = 1$$ and the given point is $$\Bigl(\dfrac1{\sqrt2}, \dfrac1{\sqrt2}\Bigr)$$, so the tangent is

$$x\Bigl(\dfrac1{\sqrt2}\Bigr) + y\Bigl(\dfrac1{\sqrt2}\Bigr) = 1.$$

Multiplying by $$\sqrt2$$ gives $$x + y = \sqrt2$$, which we rewrite in slope-intercept form:

$$y = -x + \sqrt2.$$

Thus the slope of $$L_1$$ is $$m_1 = -1$$.

The desired tangent $$y = mx + c$$ must be perpendicular to $$L_1$$, and for two lines to be perpendicular, the product of their slopes is $$-1$$. Therefore we have

$$m \cdot (-1) = -1 \;\;\Longrightarrow\;\; m = 1.$$

Hence the required tangent has the form $$y = x + c$$.

Next we impose the condition that this line is tangent to the circle $$(x - 3)^2 + y^2 = 1$$. To apply the tangency condition, we substitute $$y = x + c$$ into the circle’s equation and require that the resulting quadratic in $$x$$ have exactly one real root; equivalently, its discriminant must be zero.

Substituting, we obtain

$$\bigl(x - 3\bigr)^2 + \bigl(x + c\bigr)^2 = 1.$$

Expanding each square gives

$$(x^2 - 6x + 9) + (x^2 + 2cx + c^2) = 1.$$

Combining like terms yields

$$2x^2 + (2c - 6)x + (c^2 + 9) = 1.$$

Bringing the 1 to the left side, we have

$$2x^2 + (2c - 6)x + (c^2 + 9 - 1) = 0,$$

which simplifies to

$$2x^2 + (2c - 6)x + (c^2 + 8) = 0.$$

This is a quadratic equation in $$x$$ of the form $$Ax^2 + Bx + C = 0$$ with coefficients $$A = 2$$, $$B = 2c - 6$$ and $$C = c^2 + 8$$. For tangency we set the discriminant $$\Delta = B^2 - 4AC$$ equal to zero. Thus

$$\bigl(2c - 6\bigr)^2 - 4 \cdot 2 \cdot \bigl(c^2 + 8\bigr) = 0.$$

First, compute $$\bigl(2c - 6\bigr)^2$$:

$$\bigl(2c - 6\bigr)^2 = 4(c - 3)^2 = 4(c^2 - 6c + 9) = 4c^2 - 24c + 36.$$

Next, compute $$4 \cdot 2 \cdot \bigl(c^2 + 8\bigr)$$:

$$4 \cdot 2 \cdot \bigl(c^2 + 8\bigr) = 8c^2 + 64.$$

Setting the discriminant to zero gives

$$4c^2 - 24c + 36 - \bigl(8c^2 + 64\bigr) = 0.$$

Simplifying term by term:

$$4c^2 - 24c + 36 - 8c^2 - 64 = 0$$

$$-4c^2 - 24c - 28 = 0.$$

Multiplying both sides by $$-1$$ to make the leading coefficient positive, we obtain

$$4c^2 + 24c + 28 = 0.$$

Finally, divide every term by $$4$$ to reduce the equation to a monic quadratic:

$$c^2 + 6c + 7 = 0.$$

This matches Option C in the list. Hence, the correct answer is Option C.