Circle

Concept:Find the coordinates of points $P$, $Q$, and $R$ using the given slopes of tangents, then compute the circumradius of $\triangle PQR$.Explanation:For the parabola $y = x^2$, $\frac{dy}{dx} = 2x$.Given slope $4$, so $2x = 4 \Rightarrow x = 2$.Then $y = 2^2 = 4$, so $P(2, 4)$.For the circle $x^2 + y^2 = 2$, differentiate: $2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.Given slope $-1$, $-\frac{x}{y} = -1 \Rightarrow x = y$.$Q$ is in the first quadrant: $x^2 + x^2 = 2 \Rightarrow x = 1$, $y = 1$, so $Q(1, 1)$.For the ellipse $x^2 + 4y^2 = 8$, differentiate: $2x + 8y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{4y}$.Given slope $-\frac12$, $-\frac{x}{4y} = -\frac12 \Rightarrow x = 2y$.$R$ in first quadrant: $(2y)^2 + 4y^2 = 8 \Rightarrow 8y^2 = 8 \Rightarrow y = 1$, $x = 2$, so $R(2, 1)$.Now, $\triangle PQR$ is right-angled (verify using slopes or dot product).Length $PQ = \sqrt{(2-1)^2 + (4-1)^2} = \sqrt{10}$.The circumradius of a right triangle equals half the hypotenuse, so the required radius is $\frac{\sqrt{10}}{2} = \sqrt{\frac{5}{2}}$.Answer:$\sqrt{\dfrac{5}{2}}$ (Option C).