Let $$A$$ be a point on the $$x$$-axis. Common tangents are drawn from $$A$$ to the curves $$x^2 + y^2 = 8$$ and $$y^2 = 16x$$. If one of these tangents touches the two curves at $$Q$$ and $$R$$, then $$(QR)^2$$ is equal to

We need to find $$(QR)^2$$ where $$Q$$ and $$R$$ are the points where a common tangent from a point $$A$$ on the $$x$$-axis touches the circle $$x^2 + y^2 = 8$$ and the parabola $$y^2 = 16x$$ respectively.

A tangent to the parabola $$y^2 = 16x$$ (where $$a = 4$$) has the form $$y = mx + \dfrac{4}{m}$$ and for it to also be tangent to the circle $$x^2 + y^2 = 8$$ (radius $$= 2\sqrt{2}$$), the distance from the origin to the line must equal $$2\sqrt{2}$$:

$$\dfrac{|4/m|}{\sqrt{1 + m^2}} = 2\sqrt{2}$$

Squaring gives $$\dfrac{16}{m^2(1 + m^2)} = 8$$, which leads to $$16 = 8m^2(1 + m^2) \Rightarrow m^4 + m^2 - 2 = 0$$ and hence $$(m^2 + 2)(m^2 - 1) = 0 \Rightarrow m^2 = 1 \Rightarrow m = \pm 1$$.

Taking $$m = 1$$, the tangent is $$y = x + 4$$. Substituting into $$x^2 + y^2 = 8$$ gives $$x^2 + (x+4)^2 = 8 \Rightarrow 2x^2 + 8x + 8 = 0 \Rightarrow (x+2)^2 = 0$$, so $$x = -2, y = 2$$ and thus $$Q = (-2, 2)$$.

Substituting $$y = x + 4$$ into $$y^2 = 16x$$ yields $$(x+4)^2 = 16x \Rightarrow x^2 - 8x + 16 = 0 \Rightarrow (x-4)^2 = 0$$, so $$x = 4, y = 8$$ and hence $$R = (4, 8)$$.

Finally, $$(QR)^2 = (4-(-2))^2 + (8-2)^2 = 36 + 36 = 72$$, so the answer is $$(QR)^2 = 72$$, which corresponds to Option D.