If the tangent at (1, 7) to the curve $$x^2 = y - 6$$ touches the circle $$x^2 + y^2 + 16x + 12y + c = 0$$ then the value of c is:

We have the parabola $$x^{2}=y-6$$ and the given point $$(1,\,7)$$ lies on it because $$1^{2}=1=7-6$$.

To find the tangent at this point, we first find the slope. Differentiating the relation $$x^{2}=y-6$$ with respect to $$x$$ gives the formula $$\dfrac{dy}{dx}=2x$$.

Substituting $$x=1$$ gives the slope $$m=2\cdot1=2$$.

Using the point-slope form of a straight line, $$y-y_{1}=m(x-x_{1})$$, and substituting $$(x_{1},y_{1})=(1,7)$$ and $$m=2$$, we get

$$y-7=2(x-1).$$

Simplifying,

$$y-7=2x-2 \;\Longrightarrow\; y=2x+5.$$

Thus the tangent line is $$y=2x+5$$, which can also be written in the general form $$2x-y+5=0.$$

Next, we consider the circle $$x^{2}+y^{2}+16x+12y+c=0.$$ Writing $$x^{2}+16x$$ and $$y^{2}+12y$$ as perfect squares,

$$\bigl(x^{2}+16x+64\bigr)+\bigl(y^{2}+12y+36\bigr)= -c+64+36,$$

so $$\bigl(x+8\bigr)^{2}+\bigl(y+6\bigr)^{2}=100-c.$$

Hence the centre of the circle is $$(-8,\,-6)$$ and the radius is $$R=\sqrt{100-c}.$$

The condition for the line $$2x-y+5=0$$ to touch the circle is that the perpendicular distance from the centre to the line equals the radius. For a line $$Ax+By+C=0$$, the distance from a point $$(x_{0},y_{0})$$ is given by

$$d=\dfrac{\lvert Ax_{0}+By_{0}+C\rvert}{\sqrt{A^{2}+B^{2}}}.$$

Here, $$A=2,\;B=-1,\;C=5,\;(x_{0},y_{0})=(-8,-6).$$ Substituting,

$$\lvert 2(-8)+(-1)(-6)+5\rvert =\lvert -16+6+5\rvert=\lvert -5\rvert=5.$$

The denominator is $$\sqrt{A^{2}+B^{2}}=\sqrt{2^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}.$$

So the distance is

$$d=\dfrac{5}{\sqrt{5}}=\sqrt{5}.$$

Setting this equal to the radius,

$$\sqrt{5}=\sqrt{100-c}\;\Longrightarrow\;5=100-c\;\Longrightarrow\;c=95.$$

Hence, the correct answer is Option A.