Let P be the point on the parabola y = x^2 such that the slope of...

For the point $P$ on the parabola $y = x^2$, the slope of the tangent is $\dfrac{dy}{dx} = 2x$.
Given $2x = 4 \Rightarrow x = 2$.
Substituting $x = 2$ in $y = x^2$, we get $y = 4$.
Thus, the coordinates of $P$ are $(2, 4)$.

For the point $Q$ on the circle $x^2 + y^2 = 2$, differentiating with respect to $x$ gives $2x + 2y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{y}$.
Given $-\dfrac{x}{y} = -1 \Rightarrow x = y$.
Since $Q$ lies in the first quadrant, substituting $x = y$ in $x^2 + y^2 = 2$ gives $2x^2 = 2 \Rightarrow x = 1, y = 1$.
Thus, the coordinates of $Q$ are $(1, 1)$.

For the point $R$ on the ellipse $x^2 + 4y^2 = 8$, differentiating with respect to $x$ gives $2x + 8y \dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x}{4y}$.
Given $-\dfrac{x}{4y} = -\dfrac{1}{2} \Rightarrow x = 2y$.
Since $R$ lies in the first quadrant, substituting $x = 2y$ in $x^2 + 4y^2 = 8$ gives $4y^2 + 4y^2 = 8 \Rightarrow 8y^2 = 8 \Rightarrow y = 1, x = 2$.
Thus, the coordinates of $R$ are $(2, 1)$.

The points are $P(2, 4)$, $Q(1, 1)$, and $R(2, 1)$.
The line segment $PR$ lies on the vertical line $x = 2$, and the line segment $QR$ lies on the horizontal line $y = 1$.
Since $PR$ is perpendicular to $QR$, the angle $\angle PRQ = 90^{\circ}$.
Therefore, the triangle $PQR$ is a right-angled triangle with the hypotenuse $PQ$.
The circumcircle of $\triangle PQR$ has $PQ$ as its diameter.
The length of the diameter is $PQ = \sqrt{(2 - 1)^2 + (4 - 1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}$.
The radius of the circle is $\dfrac{PQ}{2} = \dfrac{\sqrt{10}}{2} = \sqrt{\dfrac{5}{2}}$.

Answer: $\sqrt{\dfrac{5}{2}}$