The radius of the smallest circle which touches the parabolas $$y = x^2 + 2$$ and $$x = y^2 + 2$$ is

The two parabolas are mirror images of each other across the line $$y = x$$, because inter-changing $$x$$ and $$y$$ in $$y = x^{2}+2$$ gives $$x = y^{2}+2$$. Hence the centre of the smallest circle tangent to both parabolas must lie on the line of symmetry $$y = x$$.
Let the centre be $$C(a,a)$$ and the required radius be $$r$$.

Take an arbitrary point $$P(x,\,x^{2}+2)$$ on the first parabola $$y = x^{2}+2$$.
The squared distance between $$C(a,a)$$ and $$P(x,x^{2}+2)$$ is

$$D^{2}(x,a)= (x-a)^{2}+\left(x^{2}+2-a\right)^{2}\quad -(1)$$

For the circle to be tangent, $$P$$ must be the point on the parabola that is closest to the centre. Thus, for a fixed $$a$$ we first minimise $$D^{2}(x,a)$$ with respect to $$x$$.

Differentiate $$D^{2}$$ w.r.t. $$x$$ and set the result to zero:

$$\frac{\partial D^{2}}{\partial x}=2(x-a)+2\!\left(x^{2}+2-a\right)(2x)=0$$
$$\Longrightarrow\;(x-a)+2x\!\left(x^{2}+2-a\right)=0\quad -(2)$$

Next, we must choose that particular centre $$C(a,a)$$ which makes this minimal distance itself as small as possible. Therefore we also differentiate $$D^{2}(x,a)$$ with respect to $$a$$ and set it to zero:

$$\frac{\partial D^{2}}{\partial a}=-2(x-a)-2\!\left(x^{2}+2-a\right)=0$$
$$\Longrightarrow\;(a-x)+(a-x^{2}-2)=0$$
$$\Longrightarrow\;2a=x+x^{2}+2\quad -(3)$$

Solve equations $$(2)$$ and $$(3)$$ simultaneously.

First write the differences that appear in $$(2)$$ using $$(3)$$:
$$x-a=\frac{-x^{2}+x-2}{2},\qquad x^{2}+2-a=\frac{x^{2}-x+2}{2}$$

Substituting these into $$(2)$$:

$$\frac{-x^{2}+x-2}{2}+2x\!\left(\frac{x^{2}-x+2}{2}\right)=0$$
$$\Longrightarrow\;-x^{2}+x-2+2x\!\left(x^{2}-x+2\right)=0$$
$$\Longrightarrow\;2x^{3}-3x^{2}+5x-2=0$$

Using the Rational Root Theorem, $$x=\tfrac12$$ is a root; dividing out gives
$$(x-\tfrac12)(2x^{2}-2x+4)=0$$
The quadratic factor has negative discriminant, so the only real solution is

$$x=\frac12$$

Insert $$x=\frac12$$ into $$(3)$$ to find the centre:

$$2a=\frac12+\left(\frac12\right)^{2}+2=\frac12+\frac14+2=\frac{11}{4}$$
$$\Longrightarrow\;a=\frac{11}{8}$$

Hence the centre of the required circle is
$$C\!\left(\frac{11}{8},\frac{11}{8}\right)$$

Now compute the radius. Using $$x=\tfrac12$$ and $$a=\tfrac{11}{8}$$ in $$(1)$$:

$$x-a=\frac12-\frac{11}{8}=-\frac78,\qquad x^{2}+2-a=\frac14+2-\frac{11}{8}=+\frac78$$
$$r^{2}=(x-a)^{2}+\left(x^{2}+2-a\right)^{2}=\left(\frac78\right)^{2}+\left(\frac78\right)^{2}=\frac{49}{32}$$
$$\Longrightarrow\;r=\sqrt{\frac{49}{32}}=\frac{7}{4\sqrt2}=\frac{7\sqrt2}{8}$$

The radius of the smallest circle touching both parabolas is therefore
$$\boxed{\dfrac{7\sqrt{2}}{8}}$$

Hence, the correct option is Option D.